# Incircle and circumcircle in triangle

Let $$c_1$$ be the incircle of triangle $$ABC$$. $$c_1$$ intersects the triangle $$ABC$$ in points $$D, E$$ and $$F$$. Let $$c_2$$ be another circle intersecting the points $$A$$ and $$B$$ (with points $$D, E$$ inside of $$c_2$$). Further, let line $$DE$$ intersect $$c_2$$ in points $$P$$ and $$Q$$. Prove that the circumcircle of $$PQF$$ intersects the side $$AB$$ in the center (marked X)?

How can I go about? (Poking around, I tried to apply the tangent secant theorem to a circle defined by points $$B, E$$ and $$X$$. The angle $$XBE$$ would be subtended by line $$EX$$ and angle $$AEX$$ would equal angle $$XBE$$. But this did not get me any further).

Suggested approach: Prove that $$TF \times TX = TB \times TA$$, where lines $$DE$$ and $$AB$$ intersect at point $$T$$. (In particular, independent of the $$P,Q$$, as suggested by the problem.)

Corollary: It follows that $$TF \times TX = TB \times TA = TP \times TQ$$ and thus $$F,X,P,Q$$ are concyclic as desired.

Proof of approach: One way to prove the equation is by side length chasing. Apply Menelaus on triangle $$ABC$$ to transversal $$TDE$$ to obtain $$TA/TB$$ and hence $$TA, TB$$. Then we can find $$TF, TX$$ and multiply it out.

Details of side length chasing

$$\frac{AT}{TB} \times \frac{BD}{DC} \times \frac{CE}{EA} = 1$$
$$\frac{AT} {TB} = \frac{ EA}{BD} = \frac{c+b-a}{c+a-b}$$
$$AT - TB = c \Rightarrow AT = \frac{ c (c+b-a) } { 2(b-a) } , TB = \frac{ c(c+a - b ) } { 2(b-a)}$$
$$TF = TB + BF = \frac{ c(c+a - b ) } { 2(b-a)} + \frac{c+a - b}{2} = \frac{ ((c-b+a)(c+a-b) } { 2(b-a)}$$
$$TX = TB + BX = \frac{ c(c+a - b ) } { 2(b-a)} + \frac{c}{2} = \frac{ c(c)}{ 2(b-a)}$$

Now, we multiply these terms to show that $$TA \times TB = \frac{ c^2 (c+b-a)(c+a-b) } { 4(b-a)^2} = TX \times TF$$

Additional observations, which I couldn't use directly

$$A, F, T, B$$ are harmonic conjugates. This can be shown either from
1) $$\frac{ TA}{TB} = \frac{EA}{BD} = \frac{AF}{FB}$$, or also from
2) Lines $$AD, BE, CF$$ are concurrent (at the Gergonne point)

• Clever - thanks. – Parzifal Oct 1 at 5:39
• Could you explain your suggestion in some more detail, please? It was not as clear to me as I had thought, thanks – Parzifal Oct 1 at 20:30
• Which parts are confusing to you? It's pretty straightforward (and only missing details of performing the actual calculations.) – Calvin Lin Oct 1 at 21:17
• I guess my problem is the side length chasing. As you stated, we have $TF$ x $TX$ = $TA$ x $TB$ = $TP$ x $TQ$. From Menelaus: $CE$/$EA$ x $TA$/$TB$ x $BD$/$DC$ = $1$. Let $c = AB = AX + XB, b = AC = EA + CE, a = BC = BD + DC$. From the equality of the tangent segment lengths one can deduce $EA = AF = (CA – BC +AB)/2$, $CE = DC = (CA + BC –AB)/2$, and $BF = BD = (AB – CA + BC)/2$. With $CE = DC$ Menelaus simplifies to $TA/TB x BD/EA = 1$. With this I wanted to show that $AX = XB$, but I am just turning in circles. – Parzifal Oct 3 at 15:13
• 1) Express TA, TB explicitly in terms of $a,b,$ 2) F is the tangency point. TF = TB + BF 3) X is defined as the mid point of AB. TX = TB + BX. 4) Hence, multiply out to show that $TA \times TB = TF \times TX$. – Calvin Lin Oct 3 at 17:44