Construct a right angled triangle PQR, in which ∠Q=90o, 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.
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Solution
Steps of Construction: (i) Draw a line segment QR=4.5cm (ii) At Q, draw a ray QX making an angle of 90o (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 ∠PQR which meets PR in T (vi) From T, draw perpendicular PL and PM respectively on PQ and QR. Proof: In △LTQ and △MTQ ∠TLQ=∠TMQ (Each 90o) ∠LQT=∠TQM (QT is angle bisector) QT=QT (Common) Thus, by AAS criterion of congrence △LTQ≅△MTQ And, by Corresponding Parts of Congruent Triangle (CPCT) TL=TM Therefore, T is equidistant from PQ and QR