Lesson Advanced problem on squares built externally on sides of an arbitrary triangle

Advanced problem on squares built externally on sides of an arbitrary triangle

Problem 1

  

  

Step 1.  Draw these lines, BV and CT.

         (They are shown in Figure 1 in green).

         Consider the triangles BAV and CAT.

         These triangles have congruent sides BA and TA, as well as congruent sides AV and AC.

         The angles BAV and TAV between these sides are congruent, too.

         (Each of these angles is the sum of the right angle and the common angle BAC).

         Hence, the triangles BAV and CAT are congruent.

         Then their sides BV and CT are congruent as the corresponding sides of congruent triangles.       
 
  




                    Figure 1. 

 
 
Step 2.  Straight lines BV and CT ( which you drew in part (a) ) are perpendicular.

         Indeed, the ray BV is obtained from the ray BA by rotating it clockwise by the angle ABV around the vertex B.

                 The ray TC is obtained from the ray TA by rotating it clockwise by the angle ATC around the vertex T.

         Since the rays TA and BA were originally perpendicular and since the angles ABV and ATC  are congruent (according part (a) ), 
         the statement of the step 1 is proved.

  

  

Step 3.  Consider the triangle BVC (you need to draw the segment VC, which is the diagonal 
         of the square ACUV and therefore goes through its center O2).

         The segment MO2 is the mid-line in this triangle (!).
         It is OBVIOUS (!!)

         Therefore, the segment MO2 is parallel to BV and its length is half of that of BV !!!).           


Step 4.  Consider the triangle TBC (you need to draw the segment TB, which is the diagonal 
         of the square ABST and therefore goes through its center O1).

         The segment MO1 is the mid-line in this triangle (!).
         It is OBVIOUS (!!)

         Therefore, the segment MO1 is parallel to TC and its length is half of that of TC !!!).

 
  




                    Figure 2. 

 
 
Step 5.  We are now at the finish line.

         Since TC and BV are perpendicular (step 2 in part (a)), it implies that MO1 and MO2 are perpendicular.

         Since MO2 is half of BV and since MO1 is half of TC, and since TC and BV are congruent (step 1 in part (a)), it implies that MO1 is congruent to MO2.