Question

Construct a right angled triangle $PQR$, in which $\angle Q={90}^o$, hypotenuse $PR=8cm$ and $QR=4.5cm$. Draw a bisector of angle $PQR$ and let it meet $PR$ at point $T$. Prove that $T$ is equidistant from $PQ$ and $QR$.

Answer 1

Steps of Construction:
(i) Draw a line segment $QR=4.5cm$
(ii) At $Q$, draw a ray $QX$ making an angle of ${90}^o$
(iii) With centre $R$ and radius $8cm$, draw an arc which intersects $QX$ at $P$.
(iv) Join $RP$
$△PQR$ is the required triangle
(v) Draw the bisector of $\angle PQR$ which meets $PR$ in $T$
(vi) From $T$, draw perpendicular $PL$ and $PM$ respectively on $PQ$ and $QR$.
Proof:
In $△LTQ$ and $△MTQ$
$\angle TLQ=\angle TMQ$ (Each ${90}^o$)
$\angle LQT=\angle TQM$ ($QT$ is angle bisector)
$QT=QT$ (Common)
Thus, by AAS criterion of congrence
$△LTQ\cong △MTQ$
And, by Corresponding Parts of Congruent Triangle (CPCT)
$TL=TM$
Therefore, $T$ is equidistant from $PQ$ and $QR$