Show $\measuredangle ATB=\measuredangle CTA$ in excircle configuration

Question

I recently came across the following problem in olympiad training material:

A circle $k$ is internally tangent to sides $AB, AC$ of $\Delta ABC$ and its circumcircle in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of $XY$ and $AZ$. Prove that $\measuredangle XTB=\measuredangle CTY$.

By inverting through the circle with center $A$ that is passing through $T$, I managed to transform the problem into the following, in my opinion easier, version:

The $A$-excircle of $\Delta ABC$ touches $AB, AC$ and $BC$ in points $X, Y$ and $Z$, respectively. Let $T$ be the intersection of the circumcircle of $\Delta AXY$ with $AZ$. Prove $\measuredangle ATB=\measuredangle CTA$.

However, I have not been able to solve the second version of the problem either and only got as far as $\measuredangle ATX=\measuredangle YTA$ which leaves $\measuredangle YTC=\measuredangle BTX$ to prove. Furthermore, the circle through $A, X $ and $Y $ is obviously Thales' circle with diameter $AM $ where $M $ is the center of the excircle.

Does anyone have a suggestion how to proceed or how to solve the initial problem without an inversion?

Answer 1

I'll split the solution into several small steps.

Let $ZX$ and $ZY$ intersect the circumcircle of $ABC$ again at $P$ and $Q$, respectively.

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It is enough to prove that $\triangle BXT \sim \triangle CYT$.

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Since $\angle BXT = \angle TYC$, it is enough to prove that $\dfrac{BX}{CY} = \dfrac{XT}{YT}$.

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Note that $ZA$ is a symmedian in $\triangle XYZ$. Therefore $\dfrac{XT}{YT}=\dfrac{XZ^2}{YZ^2}$.

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Prove that $XY \parallel PQ$. Conclude that $\dfrac{XZ}{YZ} = \dfrac{PZ}{QZ}$.

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Note that $XY \parallel PQ$ implies that $ZA$ is a symmedian in $\triangle ZPQ$ as well. Conclude that $\dfrac{ZP}{ZQ} = \dfrac{AP}{AQ}$.

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Prove that $P$ and $Q$ are the midpoints of arc $AB$ and $AC$, respectively.

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Prove that $\triangle ZBX \sim \triangle ZPA$ and $\triangle ZCY \sim \triangle ZQA$. Conclude that $\dfrac{BX}{XZ} = \dfrac{PA}{AZ}$ and $\dfrac{CY}{YZ}=\dfrac{QA}{AZ}$.

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Combine all ingredients.

Answer 2

Proof

Lemma Let the circle $c_1$ internally contact the circle $c_2$ at $T$, and contact the chord $AB$ of $c_2$ at $C$. Then $TC$ bisects $\angle ATB$.

We are not going to give the proof, because it's very simple, if you notice that $c_1$ and $c_2$ are homothetic with the homothetic center $T$. Now, let's turn to prove the original statement.

By the lemma, we have $ZX$ bisects $\angle AZB$ and $ZY$ bisects $\angle AZC$. Thus $$\frac{BX}{XA}=\frac{ZB}{ZA},~~~~~~\frac{CY}{YA}=\frac{ZC}{ZA}.$$ Since $XA=YA$, therefore, $$\frac{BX}{CY}=\frac{ZB}{ZC}=\frac{\sin\angle BAZ}{\sin \angle CAZ}=\frac{\dfrac{1}{2}\cdot AX\cdot AT \cdot \sin\angle BAZ }{\dfrac{1}{2} \cdot AY\cdot AT \cdot \sin\angle CAZ}=\frac{S_{\Delta AXT}}{S_{\Delta AYT}}=\frac{XT}{YT}.$$

Notice that $\angle BXT=\angle CYT$. Hence, $$\Delta BXT \sim \Delta CYT.$$ It follows that $$\angle XTB=\angle YTC,$$ which is desired.