# Lesson One property of a trapezoid

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## One property of a trapezoid

### Problem 1

In a trapezoid,  the straight line passing through the intersection point of the two lateral sides and the intersection point of its diagonals bisects each base of the trapezoid.  Prove.
 Proof Let  ABCD  be a trapezoid  (Figure 1);  G  be the                 intersection point of its lateral sides  AD  and  BC; E  and  F  be the midpoints of its bases  AB  and  DC; O  be the intersection point of its diagonals  AC  and  BD. The statement is that the four points  G,  E,  O and  F lie in one straight line. We will prove the statement in two steps. Figure 1.  To the  Problem 1 Figure 2a.  To the  Step 1 Figure 2b.  To the  Step 2

Step 1  (Figure 2a).  The three points  G,  E  and  F  lie in one straight line.  In other words,  for any trapezoid the intersection point of its lateral sides and the midpoints of its bases lie in one straight line.

It was proved in the lesson  One property of a median in a triangle  that the median of a triangle drawn from the vertex to the base bisects every segment parallel to the base and con-
necting two other sides.  In a triangle  ABG  the straight segment  GF  is the median,  while the segment  DC  is parallel to the base  AB.  Hence, the median  GF  intersects the base  DC  of the trapezoid at the midpoint of the segment  DC.  Thus the intersection points of the trapezoid's lateral sides and the midpoints of its two bases lie in one straight line.  The  Step 1  statement is proved.

Step 2  (Figure 2b).  The three points  E,  O  and  F  lie in one straight line.  In other words,  for any trapezoid the straight line passing through the midpoint of a base and the intersection point of its diagonals bisects the other base.

Indeed, let the straight line  EF  (Figure 2b)  passes through the midpoint  E  of the base  DC  and the intersection point  O  of the diagonals.  Then the triangles  OEC  and  OFA  are similar in accordance to the  AA  similarity test for triangles,  because they have two pairs of congruent angles.  Therefore,  the corresponding sides of these triangles ae proportional:

= .               (1)

Also,  the triangles  OED  and  OFB  are similar by the same reason.  Therefore,  their corresponding sides are proportional too:

= .               (2)

From proportions  (1)  and  (2)  we have

= .               (3)

Since  |EC| = |DE|,  the proportion  (3)  implies  |AF| = |FB|.  It means that the point  F  is the midpoint of the segment  AB.
Thus the midpoints of the bases of the trapezoid and the intersection point of its diagonals lie in one straight line.  The  Step 2  statement is proved.

From the proved statements of the  Step 1  and the  Step 2  we get the proof to the statement of the  Problem 1.

### Summary

For any trapezoid,  the midpoints of its bases,  the intersection point of its diagonals and the intersection point of its lateral sides lie in one straight line.

My other lessons on similar triangles in this site are
- Similar triangles,
- Similarity tests for triangles,
- Proofs of Similarity tests for triangles,
- In a triangle a straight line parallel to its side cuts off a similar triangle,
- Problems on similar triangles,
- Similarity tests for right-angled triangles,
- Problems on similarity for right-angled triangles,
- Problems on similarity for right-angled and acute triangles,
- One property of a median in a triangle   and
- Miscellaneous problems on similar triangles
under the current topic,  and
- Solved problems on similar triangles
under the topic  Geometry  of the section  Word problems.

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