Q12 of 62 Page 10

In the figure, AB is a diameter of the circle C(O, r)  chord CD is equal to the radius OC. AC and BD when produced meet at P. Prove that ∠APB = 60°.

Given: Circle C(O, r) with diameter AB and chord CD, such that OA = OC = CD.
AC and BD when produced meet at P.
To Prove:  ∠APB = 60°
Construction:Join OD, AD.
 
Proof: In ΔOCD,
OC = OD = CD ∴ ∠COD = 60°(Angle of a equilateral triangle)
As ∠CAD = ∠COD
=× 60° = 30°
Now, ∠ADP = ∠ADB = 90° (Angle in a semi-circle)
Now in ΔADP, ∠ADP + ∠DPA + ∠DAP = 180° ⇒ 90° + ∠DPA + 30° = 180°
Hence ∠DPA = 60° ⇒ ∠APB = 60°

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10 Prove that a circle drawn on one of the equal sides of an isosceles triangle as diameter bisects the base of the triangle. 11  Two circles are drawn on the sides AB and AC of DABC as diameters the circles intersect each other at a point D.
 13 Prove that the bisectors of the angles, formed by producing opposite sides (which are not parallel) of a cyclic quadrilateral, intersect each other at right-angle.
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