Question

in the given circle with centre O, AB is a chord if X is the midpoint of AB show that OX is perpendicular to AB​

Answer 1

proof below

Given : In the figure , O is center of the circle of the chord , And AB is chord . x is a point on AB such that AX=XB , We need to prove that , OX⊥AB

Construction : Join OA and OB

In ΔOAX and ΔOBX

OA=OB [raddi of the same circle]

OX=OX [common]

AX=XB [Given]

by SSS criteria

ΔOAX≅ΔOBX

The corresponding parts of the congruent triangle are congruent .

∴∠OXA=∠OXB [by c.p.c.t]

But∠OXA+∠OXB=180 degree

[linear pair]

∴∠OXA=∠OXB=90 degree

Hence ,OX ⊥ AB

hence proved!!!!

Answer 2

Answer:

In △s OAP and OBP,

OP=OP (Common)

OA=OB (Radius of circle)

PA=PB (Given)

Thus, △OPA≅△OPB (SAS rule)

Hence, ∠OPA=∠OPB=x (By cpct)

Sum of angles, ∠OPA+∠OPB=180o (Angles on a straight line)

x+x=180o

x=90∘

Thus, ∠OPA=∠OPB=90∘ 

OP is perpendicular to AB.