angle chasing, <PQC in 75-75-30 triangle, PC=PQ (2003 Zhautykov City VII P3)

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angle chasing, <PQC in 75-75-30 triangle, PC=PQ (2003 Zhautykov City VII P3)

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Isosceles Triangle

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#1 h Aug 23, 2019, 6:28 PM • 1 Y

Y by Adventure10

In the isosceles triangle $ABC$ ( $AB = AC$ , $\angle BAC = 30 {}^\circ$ ) on the side $AB$ and the median $AD$ , respectively, the points $Q$ and $P$ are chosen so that $PC = PQ$ ( $P \ne Q$ ). Find $\angle PQC$ .

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#2 h Aug 23, 2019, 9:03 PM • 1 Y

Y by Adventure10

I think it should be $Q\ne B$ , to prevent symmetry!

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#4 h Nov 27, 2020, 7:09 PM

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Obviously, $B$ lies on circle centred at $P$ with the radius $PC$ , because $PD\perp BC$ and $D$ is the midpoint of $BC$ , hence,
$\begin{align*} \angle PQC&= \angle PCQ=\frac{180^\circ- \angle QPC}{2}\\ &= 90^\circ-\angle QBC\\ &= 90^\circ-\angle ABC\\ &=90^\circ-\frac{180^\circ- \angle BAC}{2}\\ &=90^\circ-\frac{180^\circ- 30^\circ}{2}=\boxed{15^\circ}. \end{align*}$

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