Question

Prove that in a right-angled triangle, the mid-point of the hypotenuse is equidistant from the vertices.

Answer 1

Let $ABC$ be the right triangle right angled at B and AC is the hypotenuse and P is the midpoint of hypotenuse such that $AP=CP$
$\angle ABC={90}^{\circ }$

Draw a parallel line passing from P, and parallel to BC.

By converse of midpoint therem, D is the midpoint of AB.

$\angle ADP=\angle ABC={90}^{\circ }$ [$\because DP\parallel BC$]

In triangles ADP and BDP, we have

$AD=BD$ [D is the midpoint of AB]

$\angle ADP=\angle BDP={90}^{\circ }$

$DP=DP$ [Common]

$\therefore △ADP\cong △BDP$ [by SAS congruence rule]

$AP=BP$ [CPCT]

So, $AP=CP=BP$

Hence, in a right-angled triangle, the mid-point of the hypotenuse is equidistant from the vertices.