Lesson Proofs of Similarity tests for triangles

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<H2>Proofs of Similarity tests for triangles</H2>

<BLOCKQUOTE><H3>AA-test on similarity for triangles</H3>If two angles of one triangle are congruent to two angles of the other triangle then the triangles are similar.</BLOCKQUOTE><TABLE> 
 <TR>
 <TD>
Proof

Let two triangles &nbsp;{{{DELTA}}}ABC&nbsp; and &nbsp;{{{DELTA}}}DEF&nbsp; have the congruent angles&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

LA&nbsp; and &nbsp;LD&nbsp; and the congruent angles &nbsp;LB&nbsp; and &nbsp;LE&nbsp; (Figure 1).

We need to prove that

&nbsp;&nbsp;&nbsp;1)&nbsp; the remaining angles &nbsp;LC&nbsp; and &nbsp;LF&nbsp; are congruent too, &nbsp;and

&nbsp;&nbsp;&nbsp;2) &nbsp;the corresponding sides are proportional,

in accordance with the similarity definition &nbsp;(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/Similar-triangles.lesson>Similar

triangles</A>&nbsp; in this site).

1)&nbsp; Congruency of the remaining angles &nbsp;LC&nbsp; and &nbsp;LF&nbsp; follows the fact

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;that the sum of interior angles of a triangle is equal to &nbsp;180�.

2)&nbsp; On the side &nbsp;CA&nbsp; of the angle &nbsp;LC&nbsp; construct the segment &nbsp;CD1

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;congruent to the segment &nbsp;FD&nbsp; (Figure 2)&nbsp; and draw the straight

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 400, 200,  0.5, 10.5, 0.5, 5.5, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

locate(0.9, 1.0, A),
            locate(5.0, 1.0, B),
            locate(3.9, 5.4, C),

arc (1, 1, 1.0, 1.0, 304, 360),

arc (5, 1, 1.0, 1.0, 180, 250),
            arc (5, 1, 0.8, 0.8, 180, 250),

arc (4, 5, 0.8, 0.8, 75, 129),
            arc (4, 5, 1.0, 1.0, 75, 129),
            arc (4, 5, 1.2, 1.2, 75, 129),

line( 7.0, 1.0, 10.0, 1.0), 
            line( 7.0, 1.0,  9.0, 4.0),
            line( 9.0, 4.0, 10.0, 1.0),

locate( 6.9, 1.0, D),
            locate(10.0, 1.0, E),
            locate( 8.9, 4.4, F),

arc (7, 1, 1.0, 1.0, 304, 360),

arc (10, 1, 1.0, 1.0, 180, 250),
            arc (10, 1, 0.8, 0.8, 180, 250),

arc (9, 4, 0.8, 0.8, 75, 128),
 arc (9, 4, 1.0, 1.0, 75, 128),
 arc (9, 4, 1.2, 1.2, 75, 128)
)}}} 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 1. &nbsp;Triangles &nbsp;{{{DELTA}}}ABC&nbsp; and &nbsp;{{{DELTA}}}DEF

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 200, 200,  0.5, 5.5, 0.5, 5.5, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

locate(0.9, 1.0, A),
            locate(5.0, 1.0, B),
            locate(3.9, 5.4, C),

green(line(1.75, 2, 4.75, 2)),
            locate(1.40, 2.3, D1),
            locate(4.80, 2.3, E1),

arc (1,    1, 1.0, 1.0, 304, 360),

arc (1.75, 2, 1.0, 1.0, 304, 360),

arc (4, 5, 0.8, 0.8, 75, 129),
            arc (4, 5, 1.0, 1.0, 75, 129),
            arc (4, 5, 1.2, 1.2, 75, 129),

blue(line(1.75, 2, 2.00, 1)),
            locate(1.90, 1.0, F1)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 2. To the proof of 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the AA similarity test

</TD>
 </TR>
</TABLE>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;line &nbsp;D1E1&nbsp; parallel to the side &nbsp;AB&nbsp; of the triangle &nbsp;ABC&nbsp; till the intersection with the side &nbsp;BC&nbsp; at the point &nbsp;E1.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Then the angle &nbsp;LD1&nbsp; (the angle &nbsp;LE1D1C)&nbsp; is congruent to the angle &nbsp;LD, &nbsp;because &nbsp;LD1 ~ &nbsp;LA&nbsp; as the corresponding angles, &nbsp;and LA ~ &nbsp;LD.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It implies that the triangles &nbsp;{{{DELTA}}}DEF&nbsp; and &nbsp;{{{DELTA}}}D1E1C&nbsp; are congruent, &nbsp;because they have the congruent sides &nbsp;DF&nbsp; and &nbsp;D1C&nbsp; and the congruent adjacent angles

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;LD ~ LD1&nbsp; and &nbsp;LF ~ LC.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Next, &nbsp;since the straight line &nbsp;D1E1&nbsp; is parallel to &nbsp;AB, &nbsp;it cuts off proportional segments in the angle &nbsp;LC&nbsp; sides:

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AD1)/abs(D1C)}}} = {{{abs(BE1)/abs(E1C)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It was proved in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Parallelograms/Straight-line-in-a-triangle-parallel-to-its-side-cuts-off-proportional-segments-in-two-other-sides.lesson>Straight line in a triangle parallel to its side cuts off proportional segments in two other sides</A>&nbsp; in this site. &nbsp;&nbsp;From the proportion &nbsp;(1)

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;we have

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{(abs(AD1)+abs(D1C))/abs(D1C)}}} = {{{(abs(BE1)+abs(E1C))/abs(E1C)}}}, &nbsp;&nbsp;&nbsp;&nbsp; or &nbsp;&nbsp;&nbsp;&nbsp; {{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It remains to prove that the ratio &nbsp;{{{abs(AB)/abs(D1E1)}}}&nbsp; is equal to the common value of the ratios &nbsp;{{{abs(AC)/abs(D1C)}}}&nbsp; and &nbsp;{{{abs(BC)/abs(E1C)}}}. &nbsp;To do it, &nbsp;let us draw a straight line &nbsp;D1F1&nbsp; through

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the point &nbsp;D1&nbsp; parallel to the side &nbsp;BC &nbsp;(see the &nbsp;Figure 2&nbsp; or its copy the &nbsp;Figure 2-copy, &nbsp;which is placed here for your convenience). &nbsp;In accordance with the

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;same lesson, &nbsp;we have a proportion<TABLE>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AF1)/abs(F1B)}}} = {{{abs(AD1)/abs(D1C)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;From the proportion &nbsp;(3)&nbsp; we have

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{(abs(AF1)+abs(F1B))/abs(F1B)}}} = {{{(abs(AD1)+abs(D1C))/abs(D1C)}}}, &nbsp;&nbsp;&nbsp;&nbsp; or &nbsp;&nbsp;&nbsp;&nbsp; {{{abs(AB)/abs(F1B)}}} = {{{abs(AC)/abs(D1C)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(4)

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Combining the proportions &nbsp;(2)&nbsp; and &nbsp;(4)&nbsp; you get

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}} = {{{abs(AB)/abs(F1B)}}}.

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 200, 200,  0.5, 5.5, 0.5, 5.5, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

locate(0.9, 1.0, A),
            locate(5.0, 1.0, B),
            locate(3.9, 5.4, C),

green(line(1.75, 2, 4.75, 2)),
            locate(1.40, 2.3, D1),
            locate(4.80, 2.3, E1),

arc (1,    1, 1.0, 1.0, 304, 360),

arc (1.75, 2, 1.0, 1.0, 304, 360),

arc (4, 5, 0.8, 0.8, 75, 129),
            arc (4, 5, 1.0, 1.0, 75, 129),
            arc (4, 5, 1.2, 1.2, 75, 129),

blue(line(1.75, 2, 2.00, 1)),
            locate(1.90, 1.0, F1)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 2-copy. &nbsp;To the proof of 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the AA similarity test

</TD>
 </TR>
</TABLE>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Now, &nbsp;note that the segments &nbsp;F1B&nbsp; and &nbsp;D1E1&nbsp; are congruent as the opposite sides of the parallelogram &nbsp;F1BE1D1. &nbsp;Therefore, &nbsp;the last proportion became

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}} = {{{abs(AB)/abs(D1E1)}}}.

This completes the proof of the &nbsp;AA-test on similarity for triangles.

<BLOCKQUOTE><H3>SAS-test on similarity for triangles</H3>If an angle of one triangle is congruent to the angle of the other triangle and the including sides are proportional then the triangles are similar.</BLOCKQUOTE><TABLE> 
 <TR>
 <TD>
Proof

Let two triangles &nbsp;{{{DELTA}}}ABC&nbsp; and &nbsp;{{{DELTA}}}DEF&nbsp; have the congruent angles&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

LC&nbsp; and &nbsp;LF&nbsp; (Figure 3)&nbsp; and the concluding sides are proportional:

{{{abs(AC)/abs(DF)}}} = {{{abs(BC)/abs(EF)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5)

To prove the &nbsp;SAS-similarity test, &nbsp;we need to prove these two 
statements:

&nbsp;&nbsp;&nbsp;1)&nbsp; the remaining corresponding angles are congruent: &nbsp;LA ~ LD

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;and &nbsp;LB ~ LE, &nbsp;and

&nbsp;&nbsp;&nbsp;2) &nbsp;the ratio of the remaining sides &nbsp;{{{abs(AB)/abs(DE)}}}&nbsp; is equal to the common

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;value of the ratios of the other corresponding sides:

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AC)/abs(DF)}}} = {{{abs(BC)/abs(EF)}}} = {{{abs(AB)/abs(DE)}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(6)

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 400, 200,  0.5, 10.5, 0.5, 5.5, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

locate(0.9, 1.0, A),
            locate(5.0, 1.0, B),
            locate(3.9, 5.4, C),

arc (4, 5, 0.8, 0.8, 75, 129),
            arc (4, 5, 1.0, 1.0, 75, 129),
            arc (4, 5, 1.2, 1.2, 75, 129),

line( 7.0, 1.0, 10.0, 1.0), 
            line( 7.0, 1.0,  9.0, 4.0),
            line( 9.0, 4.0, 10.0, 1.0),

locate( 6.9, 1.0, D),
            locate(10.0, 1.0, E),
            locate( 8.9, 4.4, F),

arc (9, 4, 0.8, 0.8, 75, 128),
 arc (9, 4, 1.0, 1.0, 75, 128),
 arc (9, 4, 1.2, 1.2, 75, 128)
)}}} 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 3. &nbsp;Triangles &nbsp;{{{DELTA}}}ABC&nbsp; and &nbsp;{{{DELTA}}}DEF

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 200, 200,  0.5, 5.5, 0.5, 5.5, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

locate(0.9, 1.0, A),
            locate(5.0, 1.0, B),
            locate(3.9, 5.4, C),

green(line(1.75, 2, 4.75, 2)),
            locate(1.40, 2.3, D1),
            locate(4.80, 2.3, E1),

arc (1,    1, 1.0, 1.0, 304, 360),

arc (1.75, 2, 1.0, 1.0, 304, 360),

arc (4, 5, 0.8, 0.8, 75, 129),
            arc (4, 5, 1.0, 1.0, 75, 129),
            arc (4, 5, 1.2, 1.2, 75, 129),

blue(line(1.75, 2, 2.00, 1)),
            locate(1.90, 1.0, F1)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 4. &nbsp;To the proof of 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the SAS similarity test

</TD>
 </TR>
</TABLE>

in accordance with the similarity definition &nbsp;(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/Similar-triangles.lesson>Similar triangles</A>&nbsp; in this site).

1)&nbsp; On the side &nbsp;CA&nbsp; of the angle &nbsp;LC&nbsp; construct the segment &nbsp;CD1&nbsp; congruent to the segment &nbsp;FD&nbsp; (Figure 4).

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;On the side &nbsp;CB&nbsp; of the angle &nbsp;LC&nbsp; construct the segment &nbsp;CE1&nbsp; congruent to the segment &nbsp;FE, &nbsp;so that proportional segments are posed on the corresponding sides

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the angle &nbsp;LC.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Connect the points &nbsp;D1&nbsp; and &nbsp;E1&nbsp; by the straight line segment &nbsp;D1E1. &nbsp;The triangles &nbsp;{{{DELTA}}}DEF&nbsp; and &nbsp;{{{DELTA}}}D1E1C&nbsp; are congruent according to the SAS-test of the

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;triangle congruency.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Next, &nbsp;in accordance with the &nbsp;Theorem 2&nbsp; of the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Parallelograms/Straight-line-in-a-triangle-parallel-to-its-side-cuts-off-proportional-segments-in-two-other-sides.lesson>Straight line in a triangle parallel to its side cuts off proportional segments in two other sides</A>&nbsp; in this site,

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the straight line &nbsp;D1E1&nbsp; is parallel to the triangle side &nbsp;AB. &nbsp;Hence, &nbsp;the angles &nbsp;LCAB&nbsp; and &nbsp;LCD1E1&nbsp; are congruent as the corresponding angles at the parallel

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;lines &nbsp;AB&nbsp; and &nbsp;D1E1&nbsp; and the transverse &nbsp;AC.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Thus the triangles &nbsp;{{{DELTA}}}ABC&nbsp; and &nbsp;{{{DELTA}}}DEF&nbsp; have the pair of congruent angles &nbsp;LA&nbsp; and &nbsp;LE, &nbsp;in addition to the pair of the congruent angles &nbsp;LC&nbsp; and &nbsp;LF.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It implies that the angles &nbsp;LB&nbsp; and &nbsp;LE&nbsp; of these triangles are congruent too.

2)&nbsp; It remains to prove the proportion &nbsp;(6): &nbsp;{{{abs(AC)/abs(DF)}}} = {{{abs(BC)/abs(EF)}}} = {{{abs(AB)/abs(DE)}}}&nbsp;. &nbsp;Since the triangles &nbsp;{{{DELTA}}}DEF&nbsp; and &nbsp;{{{DELTA}}}D1E1C&nbsp; are congruent, &nbsp;it is equivalent to proving the proportion

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AB)/abs(D1E1)}}} = {{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}}. &nbsp;The part of the last proportion, &nbsp;namely

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(7)

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;is the consequence of the given condition &nbsp;(5)&nbsp; and the congruency of the triangles &nbsp;{{{DELTA}}}DEF&nbsp; and &nbsp;{{{DELTA}}}D1E1C, &nbsp;so it is just proved.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;To prove the rest, &nbsp;let us draw a straight line &nbsp;D1F1&nbsp; through the point &nbsp;D1&nbsp; parallel to the side &nbsp;BC &nbsp;(see the &nbsp;Figure 4&nbsp; or its copy the &nbsp;Figure 4-copy).

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In accordance with the same lesson, &nbsp;we have a proportion<TABLE>
 <TR>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AF1)/abs(F1B)}}} = {{{abs(AD1)/abs(D1C)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(8)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;From the proportion &nbsp;(8)&nbsp; we have

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Combining the proportions &nbsp;(7)&nbsp; and &nbsp;(9)&nbsp; you get

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}} = {{{abs(AB)/abs(F1B)}}}.

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 200, 200,  0.5, 5.5, 0.5, 5.5, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

locate(0.9, 1.0, A),
            locate(5.0, 1.0, B),
            locate(3.9, 5.4, C),

green(line(1.75, 2, 4.75, 2)),
            locate(1.40, 2.3, D1),
            locate(4.80, 2.3, E1),

arc (1,    1, 1.0, 1.0, 304, 360),

arc (1.75, 2, 1.0, 1.0, 304, 360),

arc (4, 5, 0.8, 0.8, 75, 129),
            arc (4, 5, 1.0, 1.0, 75, 129),
            arc (4, 5, 1.2, 1.2, 75, 129),

blue(line(1.75, 2, 2.00, 1)),
            locate(1.90, 1.0, F1)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 4-copy. &nbsp;To the proof of 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the SAS similarity test

</TD>
 </TR>
</TABLE>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Now, &nbsp;note that the segments &nbsp;F1B&nbsp; and &nbsp;D1E1&nbsp; are congruent as the opposite sides of the parallelogram &nbsp;F1BE1D1. &nbsp;Therefore, &nbsp;the last proportion become

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}} = {{{abs(AB)/abs(D1E1)}}}.

It completes the proof of the &nbsp;SAS-test on similarity for triangles.

<BLOCKQUOTE><H3>SSS-test on similarity for triangles</H3>If three sides of one triangle are respectively proportional to the tree sides of the other triangle then the triangles are similar.</BLOCKQUOTE><TABLE> 
 <TR>
 <TD>
Proof

Let two triangles &nbsp;{{{DELTA}}}ABC&nbsp; and &nbsp;{{{DELTA}}}DEF&nbsp; (Figure 5)&nbsp; have the&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

proportional sides

{{{abs(AC)/abs(DF)}}} = {{{abs(BC)/abs(EF)}}} = {{{abs(AB)/abs(DE)}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(10)

To prove the &nbsp;SSS-similarity test, &nbsp;we need to prove that the

corresponding angles are congruent: &nbsp;LA ~ LD, &nbsp;LB ~ LE&nbsp; and

LC ~ LF&nbsp; in accordance with the similarity definition &nbsp;(see the

lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/Similar-triangles.lesson>Similar triangles</A>&nbsp; in this site).

For it, &nbsp;construct the segment &nbsp;CD1&nbsp; congruent to the segment

FD&nbsp; on the side &nbsp;CA &nbsp;of the angle &nbsp;LC&nbsp; (Figure 5).

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 400, 200,  0.5, 10.5, 0.5, 5.5, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

locate(0.9, 1.0, A),
            locate(5.0, 1.0, B),
            locate(3.9, 5.4, C),

arc (1, 1, 1.0, 1.0, 304, 360),

arc (5, 1, 1.0, 1.0, 180, 250),
            arc (5, 1, 0.8, 0.8, 180, 250),

arc (4, 5, 0.8, 0.8, 75, 129),
            arc (4, 5, 1.0, 1.0, 75, 129),
            arc (4, 5, 1.2, 1.2, 75, 129),

line( 7.0, 1.0, 10.0, 1.0), 
            line( 7.0, 1.0,  9.0, 4.0),
            line( 9.0, 4.0, 10.0, 1.0),

locate( 6.9, 1.0, D),
            locate(10.0, 1.0, E),
            locate( 8.9, 4.4, F),

arc (7, 1, 1.0, 1.0, 304, 360),

arc (10, 1, 1.0, 1.0, 180, 250),
            arc (10, 1, 0.8, 0.8, 180, 250),

arc (9, 4, 0.8, 0.8, 75, 128),
 arc (9, 4, 1.0, 1.0, 75, 128),
 arc (9, 4, 1.2, 1.2, 75, 128)
)}}} 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 5. &nbsp;Triangles &nbsp;{{{DELTA}}}ABC&nbsp; and &nbsp;{{{DELTA}}}DEF

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 200, 200,  0.5, 5.5, 0.5, 5.5, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

locate(0.9, 1.0, A),
            locate(5.0, 1.0, B),
            locate(3.9, 5.4, C),

green(line(1.75, 2, 4.75, 2)),
            locate(1.40, 2.3, D1),
            locate(4.80, 2.3, E1),

arc (1,    1, 1.0, 1.0, 304, 360),

arc (1.75, 2, 1.0, 1.0, 304, 360),

arc (4, 5, 0.8, 0.8, 75, 129),
            arc (4, 5, 1.0, 1.0, 75, 129),
            arc (4, 5, 1.2, 1.2, 75, 129),

blue(line(1.75, 2, 2.00, 1)),
            locate(1.90, 1.0, F1)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 6. &nbsp;To the proof of 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the SSS similarity test

</TD>
 </TR>
</TABLE>

Next, &nbsp;construct the segment &nbsp;CE1&nbsp; congruent to the segment &nbsp;FE&nbsp; on the side &nbsp;CB&nbsp; of the angle &nbsp;LC, &nbsp;so that proportional segments are posed on the corresponding sides of the angle &nbsp;LC. &nbsp;Connect the points &nbsp;D1&nbsp; and &nbsp;E1&nbsp; by the straight line segment &nbsp;D1E1.

Due to these constructions and the given condition &nbsp;(10), &nbsp;we have a proportion

{{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(11)

Based on the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Parallelograms/Straight-line-in-a-triangle-parallel-to-its-side-cuts-off-proportional-segments-in-two-other-sides.lesson>Straight line in a triangle parallel to its side cuts off proportional segments in two other sides</A>&nbsp; in this site, &nbsp;we can conclude that the straight line &nbsp;D1E1&nbsp; is parallel to the side &nbsp;AB, &nbsp;and, &nbsp;hence, &nbsp;the angles &nbsp;LBAC&nbsp; and LE1D1C&nbsp; are congruent: &nbsp;LBAC ~ LE1D1C, &nbsp;as well as the angles &nbsp;LABC&nbsp; and &nbsp;LD1E1C&nbsp; are congruent: &nbsp;LABC ~ LD1E1C.

So, &nbsp;as soon as we prove congruency of the triangles &nbsp;{{{DELTA}}}D1E1C&nbsp; and &nbsp;{{{DELTA}}}DEF, &nbsp;we will get the proof of similarity of the triangles &nbsp;{{{DELTA}}}ABC&nbsp; and &nbsp;{{{DELTA}}}DEF.

To get the proof, &nbsp;let us draw a straight line &nbsp;D1F1&nbsp; through the point &nbsp;D1&nbsp; parallel to the side &nbsp;BC &nbsp;(see the &nbsp;Figure 6&nbsp; or its copy the &nbsp;Figure 6-copy).

In accordance with the same lesson, &nbsp;we have a proportion<TABLE>
 <TR>
 <TD>

{{{abs(AF1)/abs(F1B)}}} = {{{abs(AD1)/abs(D1C)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(12)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

From the proportion &nbsp;(12)&nbsp; we have

{{{(abs(AF1)+abs(F1B))/abs(F1B)}}} = {{{(abs(AD1)+abs(D1C))/abs(D1C)}}}, &nbsp;&nbsp;&nbsp;&nbsp; or &nbsp;&nbsp;&nbsp;&nbsp; {{{abs(AB)/abs(F1B)}}} = {{{abs(AC)/abs(D1C)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(13)

Combining the proportions &nbsp;(11)&nbsp; and &nbsp;(13)&nbsp; you get

{{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}} = {{{abs(AB)/abs(F1B)}}}.

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 200, 200,  0.5, 5.5, 0.5, 5.5, 
            line( 1.0, 1.0, 5.0, 1.0), 
            line( 1.0, 1.0, 4.0, 5.0),
            line( 4.0, 5.0, 5.0, 1.0),

locate(0.9, 1.0, A),
            locate(5.0, 1.0, B),
            locate(3.9, 5.4, C),

green(line(1.75, 2, 4.75, 2)),
            locate(1.40, 2.3, D1),
            locate(4.80, 2.3, E1),

arc (1,    1, 1.0, 1.0, 304, 360),

arc (1.75, 2, 1.0, 1.0, 304, 360),

arc (4, 5, 0.8, 0.8, 75, 129),
            arc (4, 5, 1.0, 1.0, 75, 129),
            arc (4, 5, 1.2, 1.2, 75, 129),

blue(line(1.75, 2, 2.00, 1)),
            locate(1.90, 1.0, F1)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 6-copy. &nbsp;To the proof of 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the SSS similarity test

</TD>
 </TR>
</TABLE>

Now, &nbsp;note that the segments &nbsp;F1B&nbsp; and &nbsp;D1E1&nbsp; are congruent as the opposite sides of the parallelogram &nbsp;F1BE1D1. &nbsp;Therefore, &nbsp;the last proportion become

{{{abs(AC)/abs(D1C)}}} = {{{abs(BC)/abs(E1C)}}} = {{{abs(AB)/abs(D1E1)}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(14)

Comparing the proportions &nbsp;(10)&nbsp; and &nbsp;(14)&nbsp; and taking into account the segments congruency &nbsp;|DF| ~ |D1C|&nbsp; and &nbsp;|EF| ~ |E1C|, &nbsp;you can conclude that the segments &nbsp;|DE|&nbsp; and &nbsp;|D1E1|&nbsp; are congruent too.

Thus the triangles &nbsp;{{{DELTA}}}D1E1C&nbsp; and &nbsp;{{{DELTA}}}DEF are congruent in accordance with the &nbsp;SSS-test of the triangles congruency.

It completes the proof of the &nbsp;SSS-test on similarity for triangles.

My other lessons on similar triangles in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Similar-triangles.lesson>Similar triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Similarity-tests-for-triangles.lesson>Similarity tests for triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/In-a-triangle-a-straight-line-parallel-to-its-side-cuts-off-a-similar-triangle.lesson>In a triangle a straight line parallel to its side cuts off a similar triangle</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Problems-on-similar-triangles.lesson>Problems on similar triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Similarity-tests-for-right-angled-triangles.lesson>Similarity tests for right-angled triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Problems-on-similarity-for-right-angled-triangles.lesson>Problems on similarity for right-angled triangles</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Problems-on-similarity-for-right-angled-and-acute-triangles.lesson>Problems on similarity for right-angled and acute triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/One-property-of-a-median-in-a-triangle.lesson>One property of a median in a triangle</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/One-property-of-a-trapezoid.lesson>One property of a trapezoid</A> &nbsp;&nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Miscellaneous-problems-on-similar-triangles.lesson>Miscellaneous problems on similar triangles</A>

under the current topic, &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-similar-triangles.lesson>Solved problems on similar triangles</A>

under the topic &nbsp;Geometry&nbsp; of the section &nbsp;Word problems.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.