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

Steps of Construction:

1. Draw a line segment QR = 4.5 cm

2. At Q, draw a ray QX making an angle of 90°.

3. With centre R and radius 8 cm, draw an arc which intersects QX at P.

4. Join RP. ∆PQR is the required triangle.

5. Draw the bisector of ∠PQR which meets PR at T.

6. From T, draw perpendicular TL and TM on PQ and QR, respectively.

In ∆LTQ and ∆MTQ

∠TLQ = ∠TMQ [Each 90°]

∠LQT = ∠TQM [QT is angle bisector]

QT = QT [Common]

∴ ∆LTQ ≅ ∆MTQ [By AAS axiom]

∴ TL = TM [By C.P.C.T.]

Hence, proved that T is equidistant from PQ and QR.