Proving that if two line segments are equal iff another pair of angles are equal.

Question

Question:

Suppose that $P$ and $Q$ are points on the sides $AB$ and $AC$ respectively of $\triangle ABC.$ The perpendiculars to the sides $AB$ and $AC$ at $P$ and $Q$ respectively meet at $D,$ an interior point of $\triangle ABC$. If $M$ is the midpoint of $BC,$ prove that $PM=QM$ if and only if $\angle BDP=\angle CDQ$.

My Problem: This question is to be proved by taking one of the two given things as correct and other as false and contradict both statements. There are certain things which I think are correct for the problem but I can't prove that.
$1.$ $PQCB$ is cyclic quadrilateral.
$2.$ $M$ is center of the circle$(PQCB)$.
$3.$ $P,D,C$ and $Q,D,B$ are collinear.
I found that $\triangle PDB \sim \triangle QDC$ which gives $QD*DB=PD*DC$. This is my only progress.

Thanks for providing a solution.

Answer 1

Hint:

Let $E$ and $F$ be midpoints of $BD$ and $CD$ respectively. Then $MFDE$ is parallelogram and $PE= BD/2= MF$ and $F = CD/2 = EM$. Now $$PM = PQ\iff \angle PEM = \angle MFQ \iff \angle PED = \angle DFQ \iff ...$$

Answer 2

Here is a proof of the proposition using vectors:

1. $PM=QM$ translates to $\renewcommand{\vec}{\overrightarrow}|\vec v-\vec m|=|\vec m+\vec w|$.
2. The vector pointing from $P$ to $D$ must be in the direction of $\vec v$ rotated $90^\circ$ clockwise, so it can be written as $-\widehat{\vec v}\cdot p$ for some real number $p$.
3. Similarly, the vector from $Q$ to $D$ can be written as $\widehat{\vec w}\cdot q$ for some $q$.
4. Now $\Delta BPD$ and $\Delta CQD$ are similar triangles if and only if $p=q$. And this in turn is equivalent to $\angle BDP=\angle CDQ$.

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Combining 2-4 from above we can write: $$ \vec{BD}=\vec v-\widehat{\vec v}\cdot p=2\vec m+\vec w+\widehat{\vec w}\cdot p $$ which can be rearranged to $$ \vec v-2\vec m-\vec w=(\widehat{\vec v}+\widehat{\vec w})\cdot p $$ showing that $\vec v-2\vec m-\vec w$ is parallel to $\widehat{\vec v}+\widehat{\vec w}$. Hence it is perpendicular to $\vec v+\vec w$ which is equivalent to saying $$ (\vec v-2\vec m-\vec w)\cdot(\vec v+\vec w)=0\tag1 $$

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On the other hand, the first item read $PM=QM$ which implied $$ |\vec v-\vec m|=|\vec m+\vec w| $$ which after squaring both sides and applying dot product rules gives us $$ \vec v\cdot\vec v+\vec m\cdot\vec m-2\vec v\cdot\vec m=\vec m\cdot\vec m+\vec w\cdot\vec w+2\vec m\cdot\vec w\tag2 $$

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Finally, one can check that $(2)$ is equivalent to $(1)$, and we are done:

$\angle BDP=\angle CDQ$ is equivalent to $(1)$ which is equivalent to $(2)$ which is equivalent to $PM=QM$. Done!