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°
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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