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Circle $\odot O_1$ is tangent with circle $\odot O_2$ at $P$. Two tangent lines $AE$ and $AF$ of circle $\odot O_2$ meets circle $O_1$ at $B$, $G$ and $C$, $H$, respectively. $D$ is the in-center of $\triangle ABC$. $DP$ meets $BC$ at $I$, $EI$ meets $AO_2$ at $J$.

Here is a figure:

Figure illustrating problem

Prove:

  1. $E$, $B$, $D$, $P$ are concyclic
  2. $CJ\perp AO_2$
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1  
It's intended that no further reference beyond their definition is made to $G$ and $H$, right? –  AakashM Oct 26 '12 at 12:54
    
Yes, $G$ and $H$ are of no use in the problem statement. I don't know if it will be important to the proof or not, because I don't know the answer although I spend some days on it. –  user46090 Oct 26 '12 at 12:58

1 Answer 1

This is a solution for part 1. I haven't thought about part 2 yet.

The following text may not be very rigorous. To be more specific, the solution may rely on the picture a bit. For example, it may implicitly use the fact that points $P$, $E$ and $F$ all lie on the same side of line $BC$, simply because it looks that way on the picture. I'm afraid that if I make the solution any more rigorous, it will become absolutely incomprehensible.

Step 1. Build $P'$ and several other new points.

We will forget about point $P$ and the bigger circle $\odot O_1$ for a while, and focus on the triangle $\triangle ABC$ and the smaller circle $\odot O_2$. Consider the circumcircle of $\triangle EBD$. It intersects $\odot O_2$ at point $E$. There must be a second point where these two circles intersect. Let us denote that point by $P'$.

Clearly, $E,\,B,\,D,\,P'$ are concyclic. Our plan is simple: to prove that $P=P'$. To do this, we'll construct several more lines and points. Let us denote by $Q$ the second point where line $P'D$ intersects $\odot O_2$. Let $B'$ and $C'$ be the points of intersection of $O_2$ with $P'B$ and $P'C$ respectively. You can see all these on the figure: Additional constructions

Step 2. Explore the properties of the newly constructed objects.

The objective of this step is to prove that $BC||B'C'$. We will do this by some simple angle chasing.

First of all, since $EBDP'$ is an inscribed quadrilateral, $$ \angle EP'Q=\angle ABD = 1/2 \angle ABC. $$ It is quite easy to see that $$ \angle EP'F = 90^{\circ} - 1/2 \angle BAC = 1/2(\angle ABC + \angle ACB). $$ From this we have $$ \angle FP'D = 1/2 \angle ACB = \angle ACD, $$ so $FCDP'$ is also an inscribed quadrilateral.

So, we have two inscribed quadrilaterals $EBDP'$ and $FCDP'$. From the first one we know that $\angle BED = \angle BP'D$ and from the second that $\angle CFD = \angle CP'D$. It is obvious that $\angle BED = \angle CFD$ (because $\triangle AED = \triangle AFD$), therefore $\angle BP'D = \angle CP'D$. To put it in different letters, $\angle B'P'Q = \angle C'P'Q$. This in turn means that $\angle B'O_2Q = \angle C'O_2Q$, which means that $B'C' \perp O_2Q$. Now to prove that $B'C'||BC$, it only remains to prove that $O_2Q \perp BC$.

But this is easy. We already know that $\angle EP'Q = 1/2 \angle ABC$. Also, $\angle EP'Q = 1/2 \angle EO_2Q$. So $\angle EO_2Q = \angle ABC$. Since $O_2E \perp BA$, it means that $O_2Q \perp BC$, qed.

Step 3. Prove that $P=P'$.

Alright, we now know that $BC||B'C'$. From here it will be quite easy to prove that $P=P'$.

Since $BC||B'C'$, there exists a homothety $f$ with center $P'$ that sends line $B'C'$ to line $BC$. Clearly, $f(B')=B$ and $f(C')=C$. Let us denote $\odot O'_1 = f(\odot O_2)$. Since $\odot O'_1$ is the image of $\odot O_2$ under a homothety with center $P'$, circles $\odot O'_1$ and $\odot O_2$ are tangent at $P'$. Also, since $B'$ and $C'$ lie on $\odot O_2$, $B$ and $C$ lie on $\odot O'_1$.

Let us revise what we have established. We have two circles $\odot O_1$ and $\odot O'_1$. Both of them contain points $B$ and $C$ and both of them are tangent to $\odot O_2$ at points $P$ and $P'$ respectively. But from this it is already clear that $\odot O_1$ and $\odot O'_1$ are the same, and so are points $P$ and $P'$. Done!

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