Chapter 14, Problem 5PE

To determine

To prove: that the triangle formed by joining the midpoints of the sides of an isosceles triangle is an isosceles triangle.

Explanation of Solution

Given information:

There exist a isosceles triangle $\Delta ABC$ .

Calculation:

  

Let the midpoints of its sides $AB,BC,CA$ are $E,D,F$ respectively. Now $\overline{AB}=\overline{BC}$ and since $E,F$ are their mid points,

  $\overline{AE}=\overline{EB},\overline{AF}=\overline{FC}\Rightarrow \overline{BE}=\overline{CF}\mathrm{..............}(i)$

Now $\angle EBC=\angle EFD,BE\left|\right|DF$ implies $BDFE$ is a parallelogram. Then,

  $BE=DF\mathrm{....................}(ii)$

Similarly $CDFE$ is a parallelogram and,

  $DE=CF\mathrm{....................}(iii)$

From (i), (ii), and (iii) we get, $DE=DF$ .

Hence $\Delta DEF$ is a isosceles triangle proved.