In a triangle $ABC$ we have $|AB| = |AC|$ and $\angle BAC = \alpha$. Let $P \ne B$ be a point on $AB$ and $Q$ a point on the altitude drawn from $A$ such that $|PQ| = |QC|$. Find $ \angle QPC$.
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Tags: geometry, angles, equal segments
natmath
23.09.2020 21:41
$\frac{\alpha}{2}$
For example the case when $P=A$ and also the case where $P\to B$ (not $P=B$). I'll try to think of a full solution
natmath
23.09.2020 22:21
So draw the circle with center $Q$ and contains points $B$, $C$, and $P$. Draw the diameter of this circle that contains $P$ and call the other endpoint $N$. We know $\angle PNC=\angle PBC=90-\frac{\alpha}{2}$. Because triangle $PNC$ is a right triangle with right angle $C$, we know $\angle NPC=\angle QPC=\frac{\alpha}{2}$
hopetosuccess9102
24.09.2020 05:43
Since $Q$ is on the altitude from $A$, we know that $BQ=CQ=PQ$, so we have $\angle QPC = \angle QCP = \angle QCB =\angle QBC =x$. This means $\angle BCP = 2x$ and $\angle PQB =4x \implies \angle PBC = 90-x$. We also know that $\angle PBC =90-\frac{\alpha}{2}$, so $x=\boxed{\frac{\alpha}{2}}$.