Question

Two circles intersect each other at the points P and Q. Two straight lines through the points P and Q intersect one cirele at the points A and C respectively and the other circle at the points B and D respectively. Let us prove that AC || BD. ​

Answer 1

Answer:

Question :-

Two circle intersect each other at the points P and Q. Two staight lines through the points P and Q intersect one circle at the points A and C respectively and the other circle at the points B and D respectively. Let us prove that AC || BD.

Answer :-

Given:-

• Two circles Intersect each other at the points P and Q. AB intersects the circles at the points A and B.

We have to prove :

• AC || BD

Construction :-

• A, C; B, D and P, Q are joined.

Proof :-

ACQP is a cyclic quadrilater.

∴ ∠PAC + ∠PQC = 2 right angle.

Again, PQ meets CD at the points Q.

So, ∠PQC + ∠PQD = 2 right angle.

∴ ∠PAC + ∠PQC = ∠PQC + ∠PQD

$\Rightarrow$ ∠PAC = ∠PQD

Again, PQDB is a quadrilateral.

So, ∠PBD + ∠PQD = 2 right angle.

$\Rightarrow$ ∠PBD + ∠PAC = 2 right angle. [ $\because$ ∠PQD = ∠PAC ]

Now, As the straight line AB intersects AC and BD and join of two interior angle is 180°.

∴ AC || BD (Proved).