Question

2. Two circles intersect at points P and Q. Secants drawn through P and Q intersect the circles at points A, B and C, D respectively. Draw the figure and Prove that : seg AC || seg BD. ​

Answer 1

Answer:

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Answer 2

Given:

Two circles intersect each other at points P and Q.  

To prove: segment AC || segment BD

=> Figure are in attachment

=> Join segment PQ in figure

□ABCD is a cyclic quadrilateral.  

∴ ∠PAC = ∠PQB     - (1) [Corollary of cyclic quadrilateral theorem]

Also, □PQBD is a cyclic quadrilateral.

∴ ∠PQB + ∠PBD = 180° [Theorem of cyclic quadrilateral]  

∴ ∠PAC + ∠PBD = 180° [From (1)]

But, they are a pair of interior angles on the sarpe side of transversal CD on lines BD and AC.  

∴ segment AC || segment BD [Interior angles test]

Hence its proof segment AC || segment BD