Question

prove that the mid point of the hypotenuse of a right angled triangle is equidistant from its vertices​

Answer 1

Answer:

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

Step-by-step explanation:

Answer 2

Answer:

let p is the mid point of hypo. of the right triangle ABC, right angled at B.

Draw line parallel to BC from P meeting AB at D.

Join PB

In traingle, PAD and PBD

angled PDA=angled PDA (90°each due to of mid pt. theoram)

PD=PD (common)

AD=DB(as D is the mid pt. of AB)

PDA~= PDA by sas

PA=PB (C.P.C.T)

but

PA=PC(given as P is the mid pt.)

so,

PA=PC=PB