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)?

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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)

  • $\begingroup$ Clever - thanks. $\endgroup$ – Parzifal Oct 1 at 5:39
  • $\begingroup$ Could you explain your suggestion in some more detail, please? It was not as clear to me as I had thought, thanks $\endgroup$ – Parzifal Oct 1 at 20:30
  • $\begingroup$ Which parts are confusing to you? It's pretty straightforward (and only missing details of performing the actual calculations.) $\endgroup$ – Calvin Lin Oct 1 at 21:17
  • $\begingroup$ 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. $\endgroup$ – Parzifal Oct 3 at 15:13
  • 1
    $\begingroup$ 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 $TA \times TB = TF \times TX$. $\endgroup$ – Calvin Lin Oct 3 at 17:44

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