# In the following figure, OA = OC and AB = BC. Prove that: (i) ∠AOB = 90° (ii) ∆AOD ≅ ∆COD (iii) AD = CD Proof,
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
AOB=
90 PROOVED

In triangle ABD, CBD
​AB=CB
ABO=CBO
BD{COMMON}
So triangle ABD congurent to CBD by SAS
​PROOVED

• 8
Here AOB + COB is equal to 180 bcz it forms a linear pair • 1
(1)= AO=BO AB=AO.and AB is the hypotenuse so angle AOB = 90
• 1
In triangle ABO?and triangle CBO

AB=CB

BO is common side

AO=CO

Therefore, triangle ABO?is congurent to triangle CBO?by?SSS rule of congruency.

So, ?angle ABO=angle CBO?(C.P.C.T.)

And, angle AOB = COB (C.P.C.T.)

Let angle AOB = angle COB = x.

Then, x + x = 180?

Or, 2x = 180?

Or, x = 90?

Therefore, angle AOB = 90? (Proved)

Now, In triangle ABD amd triangle BCD,

?AB=CB (Given)

angle ABO = angle CBO Or angle ABD = angle CBD

BD is common side

So triangle ABD congurent to CBD by SAS

Now, let us proceed to prove question (ii).

In triangle AOD and triangle COD,

AO = CO

OD is common side

Thus, ?AOD ? ?COD (By S-S-S rule of congruency).

?
• 1
At the ending of the answer, it should be triangle AOD os congruent to triangle COD (proved). Thats a mathematical error which gets displayed in answers every time. Please ignore it.
• 1 • 4 • 1 • 1
a) In ?AOB & ? BOC
AB=BC (given)
OA=OC (given)
OB=OB (common)
Therefore Ang.AOB=Ang.BOC (cpct)
Ang.AOB+Ang.BOC=180?
Ang. AOB+Ang.AOB= 180?
2Ang.AOB=180?
Ang.AOB=90?
b) similarly Ang.AOD =Ang.COD =90?

In ?AOD & ?COD
OA=OC (given)
Ang.AOD=Ang.COD (proved above)
OD=OD (common)
Therefore?AOD=?COD (SAS)  