7.2. Consider triangle $A B C$ with altitudes $B E$ and $C F$, where $E \in (A C)$ and $F \in (A B)$. Point $P$ lies on segment $(B E)$ such that $B P = A C$, and point $Q$ lies on the extension of segment $(C F)$, such that $F \in (C Q)$ and $C Q = A B$. Determine the measure of angle $Q A P$.
## Solution:

Let $F C \cap B E=\{O\}$. Since $m(\angle B F O)=m(\angle C E O)$ and $m(\angle F O B)=m(\angle E O C)$ (as vertical angles), it follows that $m(\angle F B O)=m(\angle E C O)$. Consider $\triangle A B P$ and $\triangle Q C A$. Since

\[
\begin{aligned}
\left.\begin{array}{rl}
A B & =Q C \\
B P & =A C \\
m(\angle A B P) & =m(\angle Q C A)
\end{array}\right\} \stackrel{\text { S.A.S }}{\Rightarrow} \triangle A B P \equiv \triangle Q C A \Rightarrow m(\angle B A P)=m(\angle C Q A) . \text { Then } m(\angle Q A P)= \\
m(\angle Q A F)+m(\angle F A P)=m(\angle Q A F)+m(\angle F Q A)=180^{\circ}-90^{\circ}=90^{\circ} .
\end{aligned}
\]

Answer: $m(\angle Q A P)=90^{\circ}$.

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