let $ABC$ be a acute angled triangle , and let $P$ and $Q$ be two points on side $BC$ . construct a point ${C}_{1}$ in such a way that the convex quad $APB{C}_{1}$ is cyclic , $Q{C}_{1}||CA$ and the point ${C}_{1}$ and $Q$ lie on the opposite side of line $AB$ .construct a point ${B}_{1}$ in such a way that the convex quad $APC{B}_{1}$ is cyclic , $Q{B}_{1}||BA$ and the point ${B}_{1}$ and $Q$ lie on the opposite side of line $AC$.

prove that the points ${B}_{1},{C}_{1},P,Q$ lie on a circle .

## cyclic quad

### Re: cyclic quad

$\text{lemma}:B_{1},A,C_{1}$ collinear.

Proof:

Suppose,they are not collinear.Extend $B_{1}A$.Let it intersects the circumcircle of $ABP$ in $C_{2}$.Now,$\angle C_{2}B_{1}Q=\angle C_{2}AB=\angle C_{2}PB$.$\therefore$ $C_{2}B_{1}QP$ is cyclic.$\therefore$ $\angle C_{2}QP=\angle C_{2}B_{1}P=\angle ACP$.$\therefore C_{2}Q$ is parallel to $AC$,which is impossible.So,$B_{1},A,C_{1}$ are collinear.

Now,$\angle C_{1}B_{1}Q=\angle C_{1}AB=\angle C_{1}PB$.$\therefore$ $B_{1},C_{1},P,Q$ lie on a circle.

[I have enjoyed the problem.Thanks for posting a nice problem]

Proof:

Suppose,they are not collinear.Extend $B_{1}A$.Let it intersects the circumcircle of $ABP$ in $C_{2}$.Now,$\angle C_{2}B_{1}Q=\angle C_{2}AB=\angle C_{2}PB$.$\therefore$ $C_{2}B_{1}QP$ is cyclic.$\therefore$ $\angle C_{2}QP=\angle C_{2}B_{1}P=\angle ACP$.$\therefore C_{2}Q$ is parallel to $AC$,which is impossible.So,$B_{1},A,C_{1}$ are collinear.

Now,$\angle C_{1}B_{1}Q=\angle C_{1}AB=\angle C_{1}PB$.$\therefore$ $B_{1},C_{1},P,Q$ lie on a circle.

[I have enjoyed the problem.Thanks for posting a nice problem]

Last edited by tanmoy on Wed Jun 24, 2015 10:49 am, edited 1 time in total.

"Questions we can't answer are far better than answers we can't question"

### Re: cyclic quad

it should be $\angle C_{2}B_{1}Q$ , not $\angle C_{2}B_{1}P$ ..... typo ?tanmoy wrote:Now,$\angle C_{2}B_{1}P=\angle C_{2}AB=\angle C_{2}PB$

### Re: cyclic quad

Yeah,it is $\angle C_{2}B_{1}Q$.So,editedTahmid wrote:it should be $\angle C_{2}B_{1}Q$ , not $\angle C_{2}B_{1}P$ ..... typo ?tanmoy wrote:Now,$\angle C_{2}B_{1}P=\angle C_{2}AB=\angle C_{2}PB$

"Questions we can't answer are far better than answers we can't question"