SD=SM

[Nadim Ul Abrar]

In triangle $ABC$ , Let $D$ be the foot of perpendicular from $A$ on $BC$ and $M$ be the midpoint of $BC$ . Points $P,Q$ are on line $AB,AC$ so that $AP=AQ$ and $PQ$ goes through $M$ . Let $S$ be the circumcenter of triangle $APQ$ .Prove that $SD=SM$

[SANZEED]

@Nadim vai,can you please provide a figure? It will really help a lot.

[Nadim Ul Abrar]

Here we go

SD+SM.PNG (30.7KiB)Viewed 3753 times

[Tahmid Hasan]

Sketch:

Let us revisit a previous BdMO problem- National Higher Secondary 2010-7.
From this problem we conclude the bisector $\ell$ of $\angle A$ is the radical axis of $\odot ABC, \odot APQ$.
So our original problem converts into proving $O'D=O'M$ where $O'$ is the midpoint of $AD'$[$D'=\ell \cap \odot ABC \neq A$.]
which can be trig bashed easily!

[Phlembac Adib Hasan]

Note: Just now I've seen that my first part is similar to Tahmid Vai's proof. Also I used Nadim Vai's figure. There is another case where $AP<AB$. It can be proved in the same way like this case.
Solution:
Using Menelaus on $ABC$ for $P,M,Q$, we find $PB=CQ$. So by sine rule $\bigcirc PBM$ and $\bigcirc CQM$ have same radius. Let the circles meet again at $E$. So $EM\perp BC$. So $\angle EBM=\angle ECM$. Now note that $\angle BPE=\angle QCE=90^{\circ}$. So $A,P,E,Q$ concyclic. Now prove $\angle EAP=\angle ECM=\angle EBM=\angle EAQ$. So $AE$ is the angle bisector of $\angle A$. As $EM\perp BC$, we can conclude $E\in \bigcirc ABC$. Also now $S$ becomes the midpoint of $AE$.

Now I'll use complex number to finish the last part.
WLOG we may assume that $\bigcirc ABC$ is the unit circle, $BC$ is parallel to the real axis and $A$ lies in first or second quadrant. So $e=-i;s=(a-i)/2;m=(b+c)/2;d=1/2(a+b+c-bc/a)$
Now finish the calculation!

[sourav das]

Let $P'$ and $Q'$ be the midpoints of $AP$ and $AQ$. Let $AM$ meet $P'Q'$ at $M'$. $AS$ meets circle $APQ$ at $Y$.
$YP=YQ$ ,$BP=CQ$ [use sine law], $\angle P=\angle Q =90$. So, $YB=YC$ and thus $YM||AD$. But a homothety with ratio 2 sends $\triangle SP'Q'$ to $\triangle YPQ$. And so, $SM'||YM||AD$ and $M'$ is the midpoint of $AM$. So, $SM'$ is the perpendicular bisector of $DM$. Therefore, $SD=SM$