Question

In the fig. AB is a diameter of circle c(o,r) and chord CD is equal to radius OD.Two chords AC and BD when produced intersect at point P.
Prove that <APB=60'.

Answer 1

plz see once ..................hope this will help u:)

      

Answer 2

Proved below.

Step-by-step explanation:

Given:

Here AB is a diameter of circle C( O, r ) and chord CD is equal to radius OD.

Construction:  

Join OC, OD and BC.

To prove: ∠APB=60°

Consider ∆COD,

OC= OD = CD = r (Radius of circle)

∴ ∆COD is equilateral.

Thus, ∠COD = 60°

$\angle CBD=\frac{1}{2}\angle COD=\frac{1}{2}\times 60$=30°

[Angle subtended by an arc on the circle is half the angle subtended by the same arc on the centre]

∠ACB = 90° [Angle in a semi circle]

∴ ∠PCB = 180° – ∠ACB

= 180° – 90°

= 90°

Consider ∠BCP,

∠PCB + ∠CBP + ∠BPC = 180°

⇒ 90° + 30° + ∠BPC = 180°

⇒ 120° + ∠BPC = 180°

⇒ ∠BPC = 180° – 120°

⇒ ∠BPC = 60°

⇒ ∠BPA = 60°

⇒ ∠APB = 60°

Hence proved.