Interior triangle bisects midpoints of exterior triangle. Find the perimeter of the exterior triangle?

Question

Hello,

To solve this I used the theorem relating parallel lines to an opposing interior triangle side to an exterior side as well as the fact that the interior is half the length of the exterior. How else could this problem be solved?

Thank you

Answer 1

You could also solve it by determining that all four triangles are congruent. Combined with the theorem you used for parallel lines opposing an interior angle, you could then determine congruent angles to show similarity of triangles (opposite angles of parallelograms, alternate angles, corresponding angles etc).

Then determine a congruent side or sides by the fact that the other three triangles share a side with the middle triangle and opposite sides of parallelograms (XY = ZC for instance) to determine congruence.

Answer 2

The sides of the larger triangle are twice the sides of the smaller one.

Thus the perimeter of the larger triangle is also twice the perimeter of the small one that is $12\times 2 = 24$

Note that the sizes indicate that all triangles are right triangles and the area of the small triangle is $6$ while the area of the large triangle is $24$ which is the same number as its perimeter.