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Cyclic Quadrilateral

ABCD is a cyc...

Question

ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AD and BC produced meet at P, show that $A P B = 60^{\circ} .$

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Solution

Given $CD = OC = Radius$

In $\triangle OCD$ :

$OD = OC = CD \Rightarrow \angle ODC \\ = \angle OCD = \angle DOC = 60^o$

In $\triangle AOD, OA = OD$ (Radius of the same circle)

Let $\angle OAD = \angle ADO = x$

Therefore $\angle AOD = 180^o -2x$ … … … (i)

In $\triangle COB, OC = OB$ (Radius of the same circle)

Let $\angle OBC = \angle OCB = y$

Therefore $\angle COB = 180^o - 2y$ … … … (ii)

Now, $180^o - 2x + 60^o + 180^o - 2y = 180^o$ (straight line angle)

$\Rightarrow x + y = 120^o$

In $\triangle APB$ :

$x + y + \angle APB = 180^o \Rightarrow \angle APB = 180^o - 120^o = 60^o$

$\\$

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