Question

$PQR$ is a triangle right angled at $P$ and $M$ is a point on $QR$ such that $PM\perp QR$. Show that $\begin{array}{rcl}P{M}^2& =& QM\times MR\end{array}$.

Answer 1

Step 1. Determine the value of $\begin{array}{rcl}& & P{M}^2.\end{array}$

It is given that $PQR$ is a triangle right angled at $P$ and $M$ is a point on $QR$ such that $PM\perp QR$.

Since $PM\perp QR$, $∆PMQ$ and $∆PMR$ are right triangles.

Apply the Pythagoras theorem in $∆PMQ$.

$\begin{array}{rcl}P{Q}^2& =& P{M}^2+Q{M}^2\\ & \Rightarrow & P{M}^2=P{Q}^2-Q{M}^2\cdots \left(1\right)\end{array}$

Apply the Pythagoras theorem in $∆PMR$.

$\begin{array}{rcl}P{R}^2& =& P{M}^2+M{R}^2\\ & \Rightarrow & P{M}^2=P{R}^2-M{R}^2\cdots \left(2\right)\end{array}$

Hence, the values are $\begin{array}{rcl}P{M}^2& =& P{Q}^2-Q{M}^2\end{array}$ and $\begin{array}{rcl}P{M}^2& =& P{R}^2-M{R}^2\end{array}$.

Step 2. Prove that $\begin{array}{rcl}P{M}^2& =& QM\times MR\end{array}$.

Apply Pythagoras theorem in $∆PQR$.

$\begin{array}{rcl}Q{R}^2& =& P{Q}^2+P{R}^2\cdots \left(3\right)\end{array}$

Add equations $\left(1\right)$ and $\left(2\right)$.

$\begin{array}{rcl}2P{M}^2& =& \left(P{Q}^2+P{R}^2\right)-Q{M}^2-M{R}^2\\ & \Rightarrow & 2P{M}^2=Q{R}^2-Q{M}^2-M{R}^2\cdots \left(4\right)\left[\mathrm{From}\left(3\right)\right]\end{array}$

Since $\begin{array}{rcl}QR& =& QM+MR\end{array}$, substitute the value of $\begin{array}{rcl}& & QR\end{array}$ into $\left(4\right)$.

$\begin{array}{rcl}2P{M}^2& =& {\left(QM+MR\right)}^2-Q{M}^2-M{R}^2\\ & \Rightarrow & 2P{M}^2=Q{M}^2+M{R}^2+2QM\times MR-Q{M}^2-M{R}^2\\ & \Rightarrow & 2P{M}^2=2QM\times MR\\ & \Rightarrow & P{M}^2=QM\times MR\end{array}$

Hence, proved that $\begin{array}{rcl}P{M}^2& =& QM\times MR\end{array}$.