Question

prove that mid point c of hypotheneus in a right angled triangle AOB is situated at equal distance from vertices O,A and B of a triangle

Answer 1

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

Answer 2

Let C be the mid point of the right triangle AOB, right angled at O.

Draw a line parallel to OC from C meeting AO at D.

Join CO.

In triangles, CAD and COD.

/_ CDA = /_ CDO ( 90° each due to converse of mid point theorem)

CD=CD (common)

AD=DO

So, triangles CAD and COD are congruent by SAS rule.

CA=CO (C.P.C.T.)

CA=CB ( C is mid point )

Hence, CA=CB=CO