Question

Let ABC be a triangle inscribed in a circle Γ. And let the center of the circle Γ be O, such that ∠AOB + ∠BOC < 180°.
Now make three lines such that one is tangent to point A, another one is tangent to point B, and another one is tangent to point C. Tangent of point B and tangent of point A will meet at point P. Tangent of point B and tangent of point C will meet at point Q.

Prove that PAOB and BOCQ are both cyclic quadilaterals​

Answer 1

To prove that quadrilateral PAOB is cyclic, we need to show that the opposite angles of this quadrilateral add up to 180 degrees. Since PA and PB are tangent to the circle, we have ∠PAB = ∠PBA, and similarly, ∠POB = ∠PBA. Therefore,

∠PAO + ∠PBO = ∠PAB + ∠POB = ∠PBA + ∠PBA = 2∠PBA

Also, since O is the center of the circle Γ, we have ∠PBA = 90° - ∠BOC/2. Substituting this into the above equation, we get:

∠PAO + ∠PBO = 2(90° - ∠BOC/2) = 180° - ∠BOC

But we are given that ∠AOB + ∠BOC < 180°, which implies that ∠PAO + ∠PBO > ∠AOB. Therefore, we have:

∠PAO + ∠PBO > ∠AOB > ∠BOC

Adding ∠BOC to both sides, we get:

∠PAO + ∠PBO + ∠BOC > ∠AOB + ∠BOC

∠PAO + ∠PBO + ∠BOC > 180°

This implies that quadrilateral PAOB is cyclic.

Similarly, to prove that quadrilateral BOCQ is cyclic, we can use the same argument as above with the roles of A and C swapped, and show that the opposite angles of quadrilateral BOCQ add up to 180 degrees. Thus, both PAOB and BOCQ are cyclic quadrilaterals.