$A B C D$ is a cyclic quadrilateral whose diagonals intersect at a point $E$ . If $∠ D B C = 70^{o}, ∠ B A C = 30^{o}$ , find $∠ B C D$ . Further, if AB= BC, find $∠ E C D$ .

70^{o} & 60^{o} .

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30^{o} & 100^{o} .

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80^{o} & 50^{o} .

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100^{o} & 30^{o} .

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Solution

The correct option is A $80^{o} & 50^{o} .$
$G i v e n - A B C D i s a c y c l i c q u a d r i l a t e r a l . I t s d i a g o n a l s A B & C D i n t e r s e c t a t E . ∠ D B C = 70^{o}, ∠ B A C = 30^{o} . T o f i n d o u t - ∠ B C D = ? A l s o, i f A B = B C t h e n ∠ E C D = ? S o l u t i o n - B C s u b t e n d s ∠ B D C & ∠ B A C t o t h e c i r c u m f e r e n c e o f t h e g i v e n c i r c l e a t D & A r e s p e c t i v e l y . ∴ ∠ B D C = ∠ B A C = 30^{o} [s i n c e t h e a n g l e s, s u b t e n d e d b y a c h o r d o f a c i r c l e t o d i f f e r e n t p o i n t s o f t h e c i r c u m f e r e c e o f t h e s a m e c i r c l e, a r e e q u a l] . ∴ ∠ B C D = 180^{o} - (∠ D B C + ∠ B D C) (a n g l e s u m p r o p e r t y o f t r i a n g l e s) ⟹ ∠ B C D = 180^{o} - (70^{o} + 30^{o}) = 80^{o} . A g a i n, i n Δ A B C w e h a v e A B = B C . i . e Δ A B C i s i s o s c e l e s w i t h A C a s b a s e . ∴ ∠ B A C = ∠ B C A = 30^{o} . S o ∠ E C D = ∠ B C D - ∠ B C A = 80^{o} - 30^{o} = 50^{o} . ∴ ∠ B C D & ∠ E C D a r e r e s p e c t i v e l y 80^{o} & 50^{o} . H e n c e, o p t i o n C i s c o r r e c t .$

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