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 $$QR$$.

Solution

## 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 $$RP$$$$\triangle PQR$$ 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 $$\triangle LTQ$$ and $$\triangle MTQ$$$$\angle TLQ=\angle TMQ$$ (Each $${ 90 }^{ o }$$)$$\angle LQT =\angle TQM$$ ($$QT$$ is angle bisector)$$QT=QT$$ (Common)Thus, by AAS criterion of congrence$$\triangle LTQ \cong \triangle MTQ$$And, by Corresponding Parts of Congruent Triangle (CPCT)$$TL=TM$$Therefore, $$T$$ is equidistant from $$PQ$$ and $$QR$$Mathematics

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