Question

Prove that the midpoint of the longest side of a right∆ is equidistant from the 3 vertices.

Answer 1

hey mate here is your ans:-

Let P be the mid point of the hypo. of the right triangle ABC, right angled at B.
Draw a line parallel to BC from P meeting AB at D.
Join PB.
in triangles,PAD and PBD,
angle PDA= angle PDB (90 each due to conv of mid point theorem)
PD=PD(common)
AD=DB( as D is mid point of AB)
so triangles PAD and PBD are congruent by SAS rule.
PA=PB(C.P.C.T.)
but
PA=PC(given as P is mid point )
So,
PA=PC=PB

hope this will help.

mark this ans. as brainlist plz.

Answer 2

Let ∆ ABC be right angled at 90°. Let D be the midpoint of the hypotenuse AC.

Draw DE || CB

Then by the converse of midpoint theorem E is the middle point of AB.

The ∆s ADE and BDE are congruent by SAS axiom. Hence AD=BD

Hence AD=BD =CD.

D the midpoint of the hypotenuse is equidistant from the vertices of the right angled triangle.