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

Explanation

Solution
Lets assume `ABC`as the right angled triangle having `90^@` at `B` and `AC` is the hypotenuse and `P` is the midpoint of hypotenuse such that

`AP=CP` and `/_ABC=90^@`
Construction
Draw a parallel line passing from `P`, and parallel to `BC`.
By converse of midpoint theorem, `D` is the midpoint of `AB.`
`/_ADP=/_ABC=90^@`[`:'DP|\ |BC`]
In triangle `ADP` and `BDP`,
we have
`AD=BD` ......[`D` is the midpoint of `AB`]
`/_ADP=/_BDP=90^@`
`DP=DP` ....... [Common]
`:./_\ADP~=/_\BDP` [by `SAS` congruence rule]
`AP=BP`[CPCT]
So,
`AP=CP=BP`
Hence Proved.