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In the following figure, OA = OC and AB = BC.

Prove that:

(i) ∠AOB = 90°

(ii) ∆AOD ≅ ∆COD

(iii) AD = CD

In triangle ABO and CBO

AB=CB

BO=BO{commom among triangle ABO , CBO

AO=CO

So, triangle ABO congurent to CBO by SSS

angle ABO=angle CBO {CPCT}

**ANGLE AOB=COB{CPCT}**

LET AOB=COB=X

X+X=180

2X=180

X=90

LET AOB=COB=X

X+X=180

2X=180

X=90

*AOB=*

*90 PROOVED*In triangle ABD, CBD

AB=CB

ABO=CBO

BD{COMMON}

So triangle ABD congurent to CBD by SAS

PROOVED

*AD=CD{CPCT}*PROOVED