This question is from exercise 1.15 of "Geometry: Euclid and Beyond".
Let two circles $\gamma$ and $\delta$ meet at a point $P$. Let the tangent to $\gamma$ at $P$ meet $\delta$ again at $B$, and let the tangent to $\delta$ at $P$ meet $\gamma$ again at $A$. Let $\theta$ be the circle through $A, B, P$. Let the tangent to $\theta$ at $P$ meet $\gamma$ and $\delta$ at $C, D$. Prove that $ PC \cong PD $.
Hint: Draw lines joining $P$ and the centres of the three circles, and look for a parallelogram.
My thought:
I have followed the hint and found the parallelogram the hint refers to. I can complete the question if I can prove it is indeed a parallelogram. But I don't see a way to prove it.
Can someone good at geometry help me out?
Edit:
I added a hand-drawn picture to illustrate my idea. In the picture, the centres of the three circles are $O_1, O_2, O_3$. So the parallelogram the hint refers to is $PO_1O_3O_2$. I don't know how to prove this is a parallelogram, but if I can prove it, I can then show that $\triangle O_1CO_3 \cong \triangle O_2DO_3$ by $SAS$ congruence. Then I'm able to show that $\triangle O_3CP \cong \triangle O_3DP$, hence solve the problem.
Can some one help to show why the quadrilateral $PO_1O_3O_2$ is a parallelogram?
Inversion solution
Make an inversion at $P$ with arbitrary radius $r$. Then Circle $\theta$ maps to line $\ell$ which is parallel to its tangent. Then images $A'$ and $B'$ of $A$ and $B$ lies on $\ell$. Similary $\varepsilon$ maps to line $a$ which is parallel to line $PB$ and $\delta $. So $A'$ and image $C'$ of $C$ lies on $a$. Since $C'$ lies on $CD$ and $B'$ lies on $PB$ we see that $A'B'PC$ is parallelogram. So $C'P = A'B'$
In the same way we deduce that $D'P=A'B'$ so $p$ halves $C'D'$ and thus it halves also $CD$.
Synthetic solution
Because of angle between tangent and a chord we have $$\angle ACP = \angle APB = \angle PDB = :\alpha$$ Similary we have $$\angle APC = \angle ABP =:\beta \;\;\;\;{\rm and}\;\;\;\;\angle BAP = \angle BPD =: \gamma$$ Since $\alpha +\beta+\gamma = 180^{\circ}$ (see angles at $P$) we have $$\triangle ACP \sim \triangle APB \sim \triangle PDB$$ so we have $${CP\over BP} = {AP\over AB} \;\;\;\;{\rm and}\;\;\;\;{DP\over AP} = {BP\over AB}$$ and thus $$ CP = {AP\cdot BP\over AB} = DP$$