April wrote: Let $ABC$ be a triangle. Circle $\omega$­ passes through points $B$ and $C.$ Circle $\omega_{1}$ is tangent internally to $\omega$­ and also to sides $AB$ and $AC$ at $T,\, P,$ and $Q,$ respectively. Let $M$ be midpoint of arc $BC\, ($containing $T)$ of ­$\omega.$ Prove that lines $PQ,\,BC,$ and $MT$ are concurrent. This is a very nice problem and now, I'll present my solution. Proof. Denote: $K=PQ\cap BC,\,K'=MT\cap BC$ Apply Menelaos's Theorem to the triangle $ABC,$ we have: $\frac{KB}{KB}\cdot \frac{QC}{QA}\cdot\frac{PA}{PB}=1\implies \frac{KB}{KC}=\frac{BP}{CQ}$ On the other hand, $M$ is the midpoint of arc $BC\,($containing $T),$ so $MT$ is the external bisector of angle $\angle BTC\implies \frac{K'B}{K'C}=\frac{TB}{TC}$ Thus, we have to pove that: $\frac{BP}{CQ}=\frac{TB}{TC}$ Denote: $E=BT\cap \omega_{1},\,F=CT\cap\omega_{1}$ We known that: $EF\parallel BC\implies\frac{BE\cdot BT}{BT^{2}}=\frac{CF\cdot CT}{CT^{2}}\quad (1)$ We also have: $\frac{BE\cdot BT}{BP^{2}}=\frac{CF\cdot CT}{CQ^{2}}\quad (2)$ From $(1)$ and $(2),$ we have: $\frac{BP}{BT}=\frac{CQ}{CT}\implies \frac{BP}{CQ}=\frac{TB}{TC}$ $\mathbb{QED}$ Remark. See: http://www.mathlinks.ro/viewtopic.php?t=145105

Here is my solution First, we construct the common targent Tx of two circles Hence, we have BMI1T is a cyclic quadrilatetal Let M is intersection of CI1 with (w). Easy to proof that MI1=MA=MX, so I1 be the incenter of triangle ABC.

I think we can improve this problem: Replace $ (O)$ by a circle which pass through $ B$ and $ C$, and T in this problem is the midpoint of arc $ BC$(not countain $ A$). The solution similar

$w_1$,B,C and T are tangent to circle $w$ (B,C and T are degenere circle) so we can do 'casey theorem' $=>|BP|.|CT|=|CQ|.|BT| $ and we have $\frac{|BP|}{|CQ|}=\frac{|BT|}{|CT|}$

Sorry for bumping, but here's my short (and quite painless) solution: Let P' and Q' be the midpoints of arc AB and AC respectively, and let I be the incenter of ABC. Let X be the intersection of BC and MT By homothety, we know that TPP' is collinear. Applying Pascal's on P'TMABC, we have that PIX collinear. Similarly we also have QIX collinear. So PQX collinear.

Beautiful! Let $X=\overline{PQ} \cap \overline{BC}$; we have (by Menelaus' Theorem) $$\frac{BX}{XC}=\frac{BP}{AP} \cdot \frac{AQ}{CQ}=\frac{BP}{CQ}.$$For circles $\omega, \omega_1,$ and the "point circle" $T$, applying the coaxial circles lemma, we get $$\frac{BT}{CT}=\frac{BP}{CQ} \Longrightarrow X \in \overline{MT},$$as desired.