Question

In the figure, AB and CD are two equal chords of the circle with centre O. Prove that : (1) ∆ AOB=COD

Answer 1

Answer:

Since AB=CD (equal chords) so, their distance from centre must le equal

So, OP=OQ

Now, In △POQ

(1) ∠OPQ=∠OQP

(2) ∠OPQ+∠OQP+∠POQ=180

0

⇒2∠OPQ+150

0

=180

0

⇒∠OPQ=15

0

Also, ∵P is midpoint of AB

OP⊥AB⇒∠APO=90

0

Now, ∠APQ=∠APO−∠OPQ=90

0

−15

0

=75

0

solution