Question

# In the following figure, AD is the diameter of the circle with centre O. Chords AB,BC and CD are equal. If ∠ DEF = 110o, calculate : (i) ∠ AEF, (ii) ∠ FAB.

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 $\angle FAB = \angle DAF + \angle OAB = 70+60=130^o$ $\\$ MathematicsConcise MathematicsStandard X

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