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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 $$QR$$
1796970_1832762_ans_735f021b7e2d4cee90fa3d68c821ecf7.png

Mathematics

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