The diagonals PR and QS of a cyclic quadrilateral PQRS intersect at X. The tangent at P is parallel to QS. Prove that PR bisects $∠$ QRS.

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Solution

$X$ must be centre of circle, $∵$ diagonals of a cyclic quadrilateral intersect at the centre of circle.
$∴ ∠ X P M = 90^{0} (∵ T a n g e n t & n o r m a l a r e a t 90^{0}) . a n d ∴ ∠ P X Q = ∠ R X Q = 90^{0} . (∵ S Q ∥ P M) . ∴ ∠ R X S = 90^{0} (S u m o f A n g l e s o n a s t r a i g h t l i n e = 180^{0}) . H e n c e, I n △ R X S l △ R X Q : \Rightarrow ∠ R X S = ∠ R X Q = 90^{0} X R = X R (c o m m o n) X S = X Q (R a d i u s) ∴ △ S R X ≅ △ R X Q (b y R . H . S .) ∴ ∠ S R X = ∠ X R Q . H e n c e, w e c a n s a y t h a t P R b i s e c t s ∠ Q R S . H e n c e, p r o v e d .$

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Q. PQRS is a quadrilateral in which diagonals PR and QS intersect at O. Prove that PQ + QR + RS + SP < 2 (PR + QS).

P Q R S

is a quadrilateral in which the diagonals intersect each other at O. Prove that

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In Fig. PQRS is a quadrilateral in which diagonals PR and QS intersect in O. [4 MARKS]

Show that:
(i) PQ + QR + RS + SP $>$ PR + QS
(ii) PQ + QR + RS + SP $<$ 2 (PR + QS)

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The diagonals PR and QS of a cyclic quadrilateral PQRS intersect at X. The tangent at P is parallel to QS. Prove that PR bisects ∠ QRS.

The diagonals PR and QS of a cyclic quadrilateral PQRS intersect at X. The tangent at P is parallel to QS. Prove that PR bisects $∠$ QRS.