Let two circles $C_1$ and $C_2$ intersect in points $A$ and $B$. The tangent to $C_1$ at $A$ intersects $C_2$ in $P$ and the line $PB$ intersects $C_1$ for the second time in $Q$ (suppose that $Q$ is outside $C_2$). The tangent to $C_2$ from $Q$ intersects $C_1$ and $C_2$ in $C$ and $D$, respectively. (The points $A$ and $D$ lie on different sides of the line $PQ$.) Show that $AD$ is the bisector of $\angle CAP$. Proposed by Iman Maghsoudi
Problem
Source: Iranian Geometry Olympiad 2016 Medium 2
Tags: geometry
iliyash
26.05.2017 14:45
Here's the full problems and solution link: http://igo-official.ir/wp-content/uploads/2016/12/igo_2016_-_problems__solutions_-_english.pdf
ThE-dArK-lOrD
26.05.2017 19:02
We have $\angle{ADC}=\angle{APD}$ and $\angle{ADP}=\angle{ABP}=180^{\circ}-\angle{ABQ}=180^{\circ}-\angle{ACQ}=\angle{ACD}$. So $\angle{CAD}=\angle{DAP}$.
smdp
17.09.2019 22:48
The proof doesn't use the condition that $AP$ is tangent to $C1$ at $A$. Therefore it isn't necessary condition.
geometry6
06.01.2021 11:54
Solution. $$\angle CAD=\angle CAB+\angle BAD=\angle CQB+\angle BDQ=\angle DBP=\angle DAP.\quad\blacksquare$$
Mahdi_Mashayekhi
27.12.2021 09:51
∠DAC = ∠DAB + ∠BAC = ∠DQB + ∠BDQ = ∠DBP = ∠DAP. we're Done.
lian_the_noob12
31.08.2023 18:14
$$\angle CAD=\angle BAC+\angle BAD=\angle CQB+\angle BDQ=\angle DBP=\angle DAP \blacksquare$$