Chapter 5, Problem 18CR

To determine

To prove that $\Delta XYZ$ is an isosceles triangle.

Explanation of Solution

Given:

  $\stackrel{-}{PX}$ and $\stackrel{-}{QY }$ are altitudes of acute $\Delta PQR$ .

  $Z$ is the midpoint of $PQ$ .

  

Proof:

Now in right triangle $\Delta PQX$ ,

  $ZX=ZP=ZQ$

  $\therefore ZQ=\frac{1}{2}PQ$ …. (i).

Now in right triangle $\Delta PQY$ ,

  $ZY=ZP=ZQ$

  $\therefore ZY=\frac{1}{2}PQ$ …..(ii).

  $\therefore$ From (i) and (ii),

  $ZX=ZY$

  $\Rightarrow \Delta XYZ$ is an isosceles triangle.

Hence proved.

Conclusion:

Therefore, $\Delta XYZ$ is an isosceles triangle.