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Sum of Opposite Angles of a Cyclic Quadrilateral

In the follow...

Question

In the following figure, AD is the diameter of the circle with centre O. Chords AB,BC and CD are equal. If $∠$ DEF = $110^{o}$ , calculate :

(i) $∠$ AEF, (ii) $∠$ FAB.

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Solution

Given $AB = BC = CD$

(i) $AOD$ is the diameter

$\angle AED = 90^o$ (angle in semicircle)

$\angle AEF + \angle AED = \angle FED$

$\angle AEF = 110-90=20^o$

(ii) Since $AB = BC = CD$ (given)

$\angle AOB = \angle BOC = \angle COD$ (equal arcs subtend equal angles at center)

We know $\angle AOD = 180^o$

$\Rightarrow \angle AOB = \angle BOC = \angle COD = 60^o$

In $\triangle AOB$

$OA = OB$ (radius of the same circle)

Therefore $\angle OAB = \angle OBA$

$\angle OAB + \angle OBA = 180-\angle AOB = 180-60 = 120^o$

$\Rightarrow \angle OAB = \angle OBA = 60^o$

$ADEF$ is a cyclic quadrilateral

$\angle DEF + \angle DAF = 180^o$ (opposite angles are supplementary)

$\angle DAF = 180-110=70^o$

Now $\angle FAB = \angle DAF + \angle OAB = 70+60=130^o$

$\\$

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