$PQR$ is a triangle right angled at $P$ and $M$ is a point on $QR$ such that $PM$ is perpendicular to $QR$. Show that $PM^2 = QM . MR$
I saw following proof somewhere. "Proof: since $PM$ is perpendicular to $QR$. Therefore $\Delta PQR$ similar to $\Delta RPM$. So, $\frac{PM}{QM} = \frac{MR}{PM} \implies PM^2 = QM . MR$". But I take $PMR$ in place of $RPM$. Then I could not get required result. Please help me.
The following dot products are zero:
$\begin{cases} \vec{PQ}\cdot\vec{PR}=0 &\quad\text{triangle rect in P}\\ \vec{PM}\cdot\vec{MQ}=\vec{PM}\cdot\vec{MR}=0&\quad(PM)\perp (QR)\text{ and }M\in(QR)\\ \end{cases}$
So we can use vector addition profitably to make these expressions appear:
$\begin{align}PM^2&=\vec{PM}\cdot\vec{PM}\\ &=(\vec{PQ}+\vec{QM})\cdot(\vec{PR}+\vec{RM})\\ &=\underbrace{\vec{PQ}\cdot\vec{PR}}_0+\vec{PQ}\cdot\vec{RM}+\vec{QM}\cdot\vec{PR}+\vec{QM}\cdot\vec{RM}\\ &=(\vec{PM}+\vec{MQ})\cdot\vec{RM}+\vec{QM}\cdot(\vec{PM}+\vec{MR})+\vec{QM} \cdot\vec{RM}\\ &=\underbrace{\vec{PM}\cdot\vec{RM}}_0+\underbrace{\vec{MQ}\cdot\vec{RM}}_{QM.MR}+\underbrace{\vec{QM}\cdot\vec{PM}}_0+\underbrace{\vec{QM}\cdot\vec{MR}}_{QM.MR}+\underbrace{\vec{QM}\cdot\vec{RM}}_{-QM.MR}\\ &=QM.MR\end{align}$
Note that we used $\ \vec x\cdot\vec y=\lVert x\lVert\times \lVert y\lVert$ when the angle between the two vectors is zero, and opposite when the angle is $180^\circ$.
$QM=MR$, $PM=PM$, and $\angle PMQ=\angle PMR=90^\circ$. Then $$\triangle PMQ\equiv\triangle PMR$$ Then it's easy to see that $$\angle QPM=\angle MPR=\frac{90^\circ}2=45^\circ$$ This means that the small triangles are isosceles, so $$QM=PM=RM$$
