Question

In the given figure, AB is diameter of circle
C(O, r). Chord CD is equal to radius OC. If AC
and BD when produced intersect at P, then prove
that (angle)APB is constant.


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Answer 1

Answer:

In△OCD

OC=OB=CD

∴∠COD=60(Anglesofequilateraltriangle)

AS∠CAD=

2

1

∠COD=

2

1

×60=30

Now∠ADP=∠ADB=90(Angleinsemicircle)

In△ADP

∠ADP+∠DPA+∠DAP=180

⇒90+∠DPA+30=180

⇒∠DPA=180−120=60=∠APB