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

Here AOB + COB is equal to 180 bcz it forms a linear pair

(1)= AO=BO AB=AO.and AB is the hypotenuse so angle AOB = 90

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

Therefore, AD = CD (Proved)


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

In triangle AOD and triangle COD,

AO = CO

OD is common side

AD = CD (we get)

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

?

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.

Here is your answer. Hope it will help you!

Please find this answer

Please find this answer

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)
c) AD=CD (cpct)

Please find this answer