Question

If the figure PQRS is a square M is the midpoint of PQ & RM $\perp$ AB. Prove that RA = RB

Answer 1

Given $PQRS$ is a square and $M$ is the midpoint of $PQ$

Also, $RM\perp AB$

In $△APM$ & $△BMQ$, we have:

$PM=MQ$ (M is the midpoint of PQ)

$\angle APM=\angle BQM=90°$

$\angle AMP=\angle BMQ$ (Vertically opposite angles)

By ASA congruence axion,

$△APM\cong △BMQ$

$\therefore AM=MB$ $\to 1$

Consider right angled $△RMA$

${RA}^2={AM}^2+{RM}^2$ (By Pythagoras theorem)

$\Rightarrow {RA}^2={MB}^2+{RM}^2\to 2$ (From $1$)

Similarly in right angled $△RMB$

${RB}^2={MB}^2+{RM}^2$ (By Pythagoras theorem) $\to 3$

From $2$ and $3$ we get,

${RA}^2={BR}^2$

$\Rightarrow AR=BR$