Lemma: Let $\Gamma$ be a circle circumscribed around the acute-angled triangle $ABC$. Let $AA_1, BB_1, CC_1$ be the altitudes of $ABC$. Let $R$ be the intersection of $BA$ and tangent to $\Gamma$ at $C$. Let $S$ be the intersection of $C$-symmedian and $AB$. Let $T$ be the intersection of $AB$ and $A_1B_1$. Claim: $(A, B; T, R) = (A, B; C_1, S)$ Proof: Note that $(A, B; S, R) = -1 \Longleftrightarrow \dfrac{\overline{AS}}{\overline{BS}}:\dfrac{\overline{AR}}{\overline{BR}} = -1$ $(A, B; C_1, T) = -1 \Longleftrightarrow \dfrac{\overline{AC_1}}{\overline{BC_1}}:\dfrac{\overline{AT}}{\overline{BT}} = -1$ $\dfrac{\overline{AT}}{\overline{BT}} \cdot \dfrac{\overline{AS}}{\overline{BS}} = \dfrac{\overline{AC_1}}{\overline{BC_1}} \cdot \dfrac{\overline{AR}}{\overline{BR}}$ $(A, B; T, R) = \dfrac{\overline{AT}}{\overline{BT}}:\dfrac{\overline{AR}}{\overline{BR}} = \dfrac{\overline{AC_1}}{\overline{BC_1}}:\dfrac{\overline{AS}}{\overline{BS}} = (A, B; C_1, S)$ Back to the main problem: Let $R$ be the intersection of $BA$ and tangent to circumcircle of $ABC$ at $C$. Let $S$ be the intersection of $C$-symmedian and $AB$. Let $T$ be the intersection of $AB$ and $A_1B_1$. Also denote $Q = AZ \cap BC, A_Z = ZC_1 \cap BC, A_P = AP \cap BC$. It is well-known that $CP$ contains $T$. $(Q, B; A_P, C) \overset{A}{=} (A, B; P, C) \overset{C}{=} (A, B; T, R) \overset{Lemma}{=} (A, B; C_1, S) \overset{Z}{=} (Q, B; A_Z, C)$ Thus $A_P = A_Z$ and lines $AP, BC, ZC_1$ are concurrent.