Chapter 9.5, Problem 23PSB

To determine

To prove:Midpoint of hypotenuse of a right triangle is equidistant from the three vertices

Explanation of Solution

Given information:Here, we have given a right-angled triangle ABC.

Proof: Let P be the midpoint of the hypotenuse. Draw a line parallel to BC from P meeting AB at D and join PB.

Now, as P is the midpoint and $DP\parallel BC$ ,

  $\Rightarrow$ D is the midpoint of AB. (By midpoint theorem) Now, In $\Delta PDA$ and $\Delta PDB$ ,

  $AD\cong DB$ (D is the midpoint)

  $PD\cong PD$ (Common)

  $\angle PDA\cong \angle PDB$ (Converse of midpoint theorem)

  $\therefore \Delta PDA\cong \Delta PDB$ (By SAS Congruency Rule)

  $\Rightarrow PA\cong PB$ (By Corresponding Parts of Congruent Triangles)

Now, as P is midpoint of AC,

  $\Rightarrow PA\cong PC$

So, $PA\cong PB\cong PC$