Dear, have a look at http://www.artofproblemsolving.com/community/c6t48f6h1290739_a_new_geometry Do you have the official solution? Sincerely Jean-Louis

I found a cute and short solution: 1- Let $R, S$ be the midpoints of the arcs $PQ$ on $\Gamma_1$ and $\Gamma_2$ respectively. 2- Let $O=SR\cap AB$ 3- It is easy to see that inversion with center at $O$ which sends $B$ to $A$, also sends $C$ to $D$ and $S$ to $R$, then $SRCP$ is cyclic. 4- $CR$, $DS$ and $PQ$ are concurrent(radical axis of three circles) 5- Conclude that the angle bisectors of angles $PDQ$ and $PCQ$ meet at a point on $PQ$, which implies the desired result.

Let's bashing! Let $AD \cap QP \cap BC=E,PQ \cap AB=M(\text{idpoint})$ Since $DP=\frac{EP*sin \angle AEM}{sin \angle AQM},CP=\frac{EP*sin \angle BEM}{sin \angle BQM}$,with $\frac{AQ}{BQ}*\frac{sin \angle AQM}{sin \angle BQM}=1$, We get $\frac{DP}{CP}=\frac{AQ*sin \angle AEM}{BQ*sin \angle BEM}$ As $\frac{DQ}{CQ}=\frac{\frac{EQ*sin \angle AEM}{sin \angle ADQ}}{\frac{EQ*sin \angle BEM}{sin \angle BCQ}}$ So it's equivalent to prove $\frac{AQ}{BQ}=\frac{sin \angle BCQ}{sin \angle ADQ}$,but $\frac{sin \angle BCQ}{sin \angle ADQ}=\frac{AP}{BP},\Leftrightarrow \frac{AQ}{BQ}=\frac{AP}{BP}$,which is obvious.Done! Your sincerely CeuAzul

Dear Mathlinkers, I research always a synthetic proof without trigonometry.. Sincerely Jean-Louis

@jayme Did you really has a solution without inversion and trigonometry?Please tell us! Your sincerely CeuAzul

You can also use inversion with center $P$ radius $PQ$ and let $X^*$ be image of $X$ under the inversion. Thus $A^*B^*D^*C^*$ is cyclic and $(O_1)^*,(O2)^*$ are $A^*D^*, B^*C^*$ resp. so $A^*D^* \cap B^*C^* = {Q}$. Since $\triangle PCQ$ and $\triangle PQC^*$ are similar we have $PC/CQ=PQ/QC^*$ , similar we get $PD/DQ=PQ/QD^*$ so it's enough to prove $QD^*=QC^* \implies A^*C^*D^*B^*$ is isosceles trapezoid and this is just angle chasing, it's enough to prove $\angle A^*C^*D^* = \angle B^*D^*C^*$ which is not hard to prove since AB is tangent to $(O_1), (O_2)$.

Dear, I haven't any proof... There for, I resarch one can be more basic... Sincerely Jean-Louis

Inversion is basic,..., basically similarity,

I found an extremely cute soln...take the point A as the center of inversion ..and the problem essentially crumbles down to the well known angle bisector theorem..

Since $AD, PQ, BC$ meet at radical centre $T$, we have $TA\cdot TD=TB\cdot TC=TP\cdot TQ$. Now consider the four sets of similarities: (1) $\triangle TAP\sim \triangle TQD$ so $\frac{1}{QD}=\frac{TP}{TD\cdot AP}$. (2) $\triangle TAQ\sim \triangle TPD$ so $PD=\frac{TP\cdot AQ}{AT}$. (3) $\triangle TBP\sim \triangle TQC$ so $\frac{1}{QC}=\frac{TP}{TC\cdot BP}$. (4) $\triangle TBQ\sim \triangle TPC$ so $PC=\frac{TP\cdot BQ}{BT}$. The first two gives $$\frac{PD}{QD}=\frac{TP^{2}\cdot AQ}{TD\cdot TA\cdot AP}=\frac{TP^{2}\cdot AQ}{TQ\cdot TP\cdot AP}=\frac{AQ\cdot TP}{TQ\cdot AP}$$ The last two gives $$\frac{CP}{CQ}=\frac{TP^{2}\cdot BQ}{TC\cdot TB\cdot BP}=\frac{TP^{2}\cdot BQ}{TQ\cdot TP\cdot BP}=\frac{BQ\cdot TP}{TQ\cdot BP}$$ But since $A,B$ are common tangency points of the two circles, it is well-known that $\frac{AQ}{AP}=\frac{BQ}{BP}$. Hence done.

Note that by Radical Axis Theorem we have $AD$ and $PQ$ and $BC$ are concurrent at $S$. $\frac{DP}{DQ} = \frac{\sin{DQP}}{\sin{DPQ}} = \frac{\sin{PAS}}{\sin{QAS}}$ and $\frac{CP}{CQ} = \frac{\sin{CQP}}{\sin{CPQ}} = \frac{\sin{PBS}}{\sin{QBS}}$ so we need to prove $\frac{AP}{AQ} = \frac{BP}{BQ}$ or $\frac{\sin{PQA}}{\sin{QAB}} = \frac{\sin{PQB}}{\sin{QBA}}$ which is true since $PQ$ bisects $AB$.

A simple but instructive exercise on radical axis. Let $M = \overline{PQ} \cap \overline{AB}$, which is the midpoint of $\overline{AB}$ by radical axis. Notice that $$\frac{PC}{QC} = \frac{QA}{PA} = \frac{AM}{MP} = \frac{BM}{MP} = \frac{QB}{PB} = \frac{PD}{QD}$$by similar triangles, so we are done.

Let $\odot O_1$ and $\odot O_2$ intersect at $P$ and $Q$, their common external tangent touches $\odot O_1$ and $\odot O_2$ at $A$ and $B$ respectively. A circle $\Gamma$ passing through $A$ and $B$ intersects $\odot O_1$, $\odot O_2$ at $D$, $C$. Prove that $\displaystyle \frac{CP}{CQ}=\frac{DP}{DQ}$