Prove that tangents and line are concurrent

Question

Consider a triangle $XYZ$ with altitudes $XA$ and $ZC$. Let the intersection between the altitude from $Y$ and the circumcircle of $XYZ$ be $B$. Let the intersection between circles $XYZ$ and $ABC$ be $P$. Prove that $YP$ and the two tangents to $XYZ$ at $X$ and $Z$ are concurrent.

Answer 1

Ok, I seriously think I deserve a lot of credit for writing this proof! I expect a lot of upvotes and a very big thank you! :D

Let the circumcircle of triangle $XYZ$ be $k_O$ with center $O$. Let the line $YO$ intersect $k_O$ for the second time at point $D$ (the first point of intersection being $Y$). Observe, $DY$ is a diameter of $k_O$. Let $H$ be the orthocenter of triangle $XYZ$ (the common intersection point of the three altitudes $XA, \, YB, \, ZC$). Let line $DH$ intersect circucircle $k_O$ at point $E$ for the second time (the first point of intersection being point $D$). Next, let the altitude $YB$ intersect the edge $XZ$ at point $N$. Let the two tangents to the circle $k_O$ at points $X$ and $Z$ intersect at point $S$ and let the circle centered at point $S$ and of radius $SX = SZ$ be denoted by $k_S$.

Observe that by construction, circle $k_S$ and $k_O$ are orthogonal.

Lemma 1. The three points $E, \, B$ and $S$ are collinear.

Proof: Let $R$ be the point of intersection of lines $BD$ and $EY$. Since $O \, \in \, DY$ i.e. $DY$ is a diameter of $k_O$ it follows that $\angle \, DEY = \angle \, DBY = 90^{\circ}$. Therefore point $H$ is an orthocenter of triangle $RDY$ and quad $BHER$ is inscribed in a circle $k_G$ with center $G$, which is the midpoint of $HR$ (because $HR$ is a diameter of $k_G$). Since $BD$ is orthogonal to $BY$ and $BY$ is orthogonal to $XZ$ (as an altitude in $XYZ$), the two lines $BD$, which is the same as $DR$, and $XZ$, which passes through $N$, are parallel to each other. Furthermore, it is a standard fact that point $N$ is the midpoint of segment $HB$. Hence, the midpoint $G$ of $HR$, which is also the center of circle $k_G$, lies on $XZ$. A direct angle chasing shows that $\angle \, GEH = 90^{\circ}$ yielding that circles $k_G$ and $k_O$ are orthogonal. Furthermore, we have shown that $G$, the center of $k_G$, lies on $XZ$ which is the radical axis of circles $k_O$ and $k_S$. Therefore, by the radical axis theorem, if a circle has a center lying on the radical axis of two other circles and is orthogonal to one of them, it is orthogonal to the other one too. Consequently, circle $k_G$ is orthogonal to circle $k_S$. Therefore, circle $k_S$ is orthogonal to both $k_G$ and $k_O$. Thus, its center $S$ lies on the radical axis of $k_G$ and $k_O$ and that radical axis is $BE$.

Denote by $\hat{k}$ the circle passing through the points $A, B$ and $C$. As stated in the problem, point $P$ is the second point of intersection of circle $\hat{k}$ with circle $k_O$, the first one being point $B$.

Lemma 2. The four lines $BP, \, XZ, \, AC$ and $EY$ intersect at a common point $Q$.

Proof: Since segment $HY$ has $90^{\circ}$ angle of view from all the points $A, C$ and $E$, the five points $A, \, H, \, C, \, E$ and $Y$ lie on a common circle $k_1$. Furthermore, for the same reason, quad $XZAC$ is inscribed in a circle $k_2$. By the radical axis theorem, the radical axes $EY, \, CA$ and $XZ$ between the corresponding pairs from the three circles $k_1, \, k_2$ and $k_O$ are concurrent and let $Q$ be their common intersection point.Next, if we apply the radical axis theorem one more time, this time to all three pairs of circles $k_1, \, \hat{k}$ and $k_O$, the corresponding three radical axes $EY, \, AC$ and $BP$ are also concurrent and since $Q$ is already the intersection point of $EY$ and $AC$, the third radical axis $BP$ also passes through $Q$.

Corollary 3. The three lines $BP, \, XZ$ and $EY$ intersect at the common point $Q$.

Draw the circle $k_{BEQ}$ passing through the three points $B, \, E$ and $Q$.

Lemma 3. Circle $k_{BEQ}$ is orthogonal to circle $k_S$.

Proof: The radical axis of $k_{BEQ}$ and $k_O$ is $BE$ and by Lemma 1 it passes through the center $S$ of circle $k_S$. Again by one of the radical axis theorems, since $S$ is on the radical axis $BE$ of $k_{BEQ}$ and $k_O$, and since circle $k_S$ is orthogonal to $k_O$, it should be also orthogonal to circle $k_{BEQ}$. (That theorem states that if a circle has a center lying on the radical axis of two other circles and is orthogonal to one of them, it is orthogonal to the other one too.)

Completing the proof: Let $k^*$ be the circle centered at point $Q$ and orthogonal to circle $k_O$. In other words, $k^*$ is centered at point $Q$ and passes through the two points of tangency of the tangent lines from $Q$ to the circle $k_O$. Then if one performs inversion with respect to $k^*$, the orthogonal circle $k_O$ is mapped to itself and since by Corollary 3 the three lines $EY, \, XZ$ and $BP$ pass through the center $Q$ of $k^*,$ point $E$ is mapped to point $Y$, point $X$ is mapped to point $Z$ and point $B$ is mapped to point $P$. The latter fact implies that circle $\hat{k}$ is orthogonal to $k^*$ so $\hat{k}$ is mapped to itself under the inversion with respect to $k^*$. Furthermore, the fact that point $X$ is mapped to point $Z$ means that circle $k_S$ is mapped to itself under inversion in $k^*$. Thus, when inverted in $k^*$, the circle $k_{BEQ}$ is mapped to the straight line $PY$ (this is so because $k_{BEW}$ passes through the center $Q$ of $k^*$, and because $B$ is mapped to $P$ and $E$ is mapped to $Y$). However, during the inversion, circle $k_S$ is mapped to itself. Therefore, since circle $k_{BEQ}$ is orthogonal to $k_S$ its image, the line $PY$, is also orthogonal to the image of $k_S$, which is $k^S,$ so $PY$ is orthogonal to $k_S$. However, a line is orthogonal to a circle if and only if it passes through the circle's center, which means that $PY$ passes through point $S$, the center of circle $k_S$. In other words, the three points $Y, \, P$ and $S$ are collinear, where $S$ is the intersection point of the tangents to the circumcircle of triangle $XYZ$ at points $X$ and $Z$.