Question

The two circles in the figure below intersect at P and Q and lines through these points meet the circles at A, B, C and D. The lines AC and BD are not parallel. Prove that if these lines are of equal length, then ABDC is a cyclic quadrilateral.

Answer 1

Extend AC and BD to join at R.
As APOB is cyclic quadrilateral,
∠BAP +∠BQP = 180°
∠BQP = 180° - ∠1
∠BQP +∠DQP = 180° (By linear pair)
180° - ∠1 +∠DQP = 180°
∠DQP = ∠ 1
Also CDQP is a cyclic quadrilateral.

∠PQD - ∠PCD = 180°
∠1 + ∠PCD = 180°
∠PCD = 180° - ∠1
So, ∠RCD=∠1....... (Linear Pair)

Similarly, ∠QDC = 180° - ∠2
So, ∠RDC=∠2..... (Linear Pair).

Now, ∠RCD= ∠A (=∠1) and ∠RDC=∠B (=∠ 2);
Hence, AB || CD
[As, corresponding angles are equal through transversal AR]

Now, In △ABR, AB || CD and CD intersects AR and BR.
By basic proportionality theorem: $\frac{AR}{AC}=\frac{BR}{BD}$
AC = BD is given.
So, AR = BR Threfore, ∠1=∠2 (angle opposite to equa sides are always equal.)
Now, ∠A+∠D=∠1+180°-∠2
=∠1+180°-∠1 (∠1=∠2)
=180°
Sum of opposite angles is 180° so quadrilateral ABDC is cyclic.