Let $ABC$ be an acute-angled triangle such that $AB<AC$. Let $D$ be the point of intersection of the perpendicular bisector of the side $BC$ with the side $AC$. Let $P$ be a point on the shorter arc $AC$ of the circumcircle of the triangle $ABC$ such that $DP \parallel BC$. Finally, let $M$ be the midpoint of the side $AB$. Prove that $\angle APD=\angle MPB$. Proposed by Dominik Burek, Poland
Problem
Source: 2019 MEMO Problem T-5
Tags: geometry, perpendicular bisector, circumcircle, memo, MEMO 2019, sinus
MarkBcc168
30.08.2019 16:27
Let $T$ be the intersection between tangents to $\odot(ABC)$ at $A,B$. Then notice that $\triangle BTA\stackrel{+}{\sim}\triangle BDC$ so $\triangle BTD\stackrel{+}{\sim}\triangle BAC$ so $\angle BTD = \angle BAC = 180^{\circ}-\angle TBC$ $\implies DT\parallel BC$ or $T\in DP$.
Now, let $DP$ intersect $\odot(ABC)$ again at $Q$. Notice that $PAQB$ is harmonic quadrilateral thus $PQ$ is symmedian while $PM$ is median of $\triangle PAB$. This implies the desired isogonality.
timon92
30.08.2019 16:30
This problem was proposed by Burii.
Steff9
01.09.2019 01:36
Let $PD\cap (ABC)=Q$. Since $P$ lies on the side bisector of $BC$ and $BC\parallel PQ$, we get $DP=DQ$. Therefore, $AD$ is median in $\triangle APQ$. Since $PQ\parallel BC$, we have that arcs $PC$ and $QB$ are equal, and therefore $\angle PAD\equiv \angle PAC=\angle QAB$. From these two points, we can conclude that $AB$ is symmedian in $\triangle APQ$. Therefore, $AQBP$ is harmonic quadrilateral. Therefore, $PQ\equiv PD$ is symmedian in $\triangle PAB$ and finally $\angle APD=\angle MPB\,\blacksquare $
bora_olmez
07.06.2021 00:28
Similar to the solutions above but not completely identical - so I guess I will post it. Let $DP \cap circ(ABC) = T$ and $T \neq P$. Then, notice that $D$ is the midpoint of $TP$ because $D$ lies on the perpendicular bisector of $BC$ and $DP$ is parallel to $BC$. Let $P_{\infty}$ be the point at infinity along $BC$ and therefore also along $DP$. Then, $$ -1 = (T,P;D,P_{\infty}) \stackrel{C}{=}(T,P;A,B)$$We therefore get that $PT$ is a symmedian in $\triangle APB$ which by the Symmedian Lemma implies that $$\angle APD = \angle APT = \angle MPB$$
lazizbek42
10.01.2022 19:09
$$\frac{sin \angle BAP}{sin \angle ABP}=\frac{sin \angle DPB}{sin \angle APD}$$$PD$ symmedian in $\triangle APB$
Jasurbek
17.01.2022 19:52
Lazizbek42 good solution I solved the same
timon92
08.03.2022 06:55
No symmedian mumbo-jumbo needed. Let $DP$ intersect the circumcircle of $ABC$ again at a point $Q$. We have $\angle ADQ = \angle ACB = \angle APB$ and $\angle AQD = \angle ABP$. Hence $\triangle AQD \sim \triangle ABP$. Therefore $\dfrac{AQ}{QD} = \dfrac{AB}{BP}$. Since $QP=2QD$ and $AB=2MB$, we get $\dfrac{AQ}{QP}=\dfrac{MB}{BP}$. Since $\angle AQP = \angle MBP$, we obtain $\triangle AQP \sim \triangle MBP$. Hence $\angle APD = \angle MPB$.
Kimchiks926
25.06.2022 00:04
Suppose $DP \cap \odot(ABC)=X$. The problem asks us to show that $PX$ is symmedian in $\triangle APB$, which is equivalent to quadrilateral $APBX$ being harmonic. Obviously by symmetry $D$ is the midpoint of the segment $XP$. Therefore: $$ -1 =(X,P; D, \infty) \overset{C}{=} (X,P;A;B) $$where we used the fact that $XP \parallel BC$. Since$ (X,P;A;B)=-1$, we indeed have that quadrilateral $APBX$ is harmonic as desired.