Question

1) Two circles intersect each other at the points P and Q. Two straight line through the point 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 1

Answer:

Given:-

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

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

➭ or, ∠PAC = ∠PQD

Again, PQDB is a quadrilateral .

So, ∠PBD + ∠PQD =2 right angle

➭ ∠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 )

Answer 2

Step-by-step explanation:

Given:-

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

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

➭ or, ∠PAC = ∠PQD

Again, PQDB is a quadrilateral .

So, ∠PBD + ∠PQD =2 right angle

➭ ∠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 )