In the given figure, two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and P, Q respectively. Prove that : seg SQ || seg RP.

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Solution

It is given that two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and P, Q.

Join MN.

Quadrilateral PRMN is a cyclic quadrilateral.

∴ ∠PRM = ∠MNQ .....(1) (Exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle)

Quadrilateral QSMN is a cyclic quadrilateral.

∴ ∠QSM = ∠MNP .....(2) (Exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle)

Adding (1) and (2), we get

∠PRM + ∠QSM = ∠MNQ + ∠MNP .....(3)

Now,

∠MNQ + ∠MNP = 180º .....(4) (Angles in linear pair)

From (3) and (4), we get

∠PRM + ∠QSM = 180º

Now, line RS is transversal to the lines PR and QS such that

∠PRS + ∠QSR = 180º

∴ seg SQ || seg RP (If the interior angles formed by a transversal of two distinct lines are supplementary, then the two lines are parallel)

Hence proved.

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Question 9
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that $∠ A C P = ∠ Q C D$

Q. Two circles with centres

M

and

N

intersect each other at

P

and

Q .

The tangents drawn from point

R

on the line

P Q

touch the circles at

S

and

T

Prove that,

R S = R T .

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