Question

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 meets PR at point T. Prove that T is equidistant from PQ and QR. 

Answer 1

 
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. ΔPQR is the required triangle.
5. Draw the bisector of ∠PQR that meets PR in T.
6. From T, draw perpendicular PL and PM respectively on PQ and QR.

Proof: In ΔLTQ and ΔMTQ 

∠TLQ = ∠TMQ  ...(Each = 90°)

∠LQT = ∠TQM  ...(QT is angle bisector)

QT = QT  ...(Common)

∴ By Angle – Angle – side criterion of congruence,

∴ ΔLTQ ≅ ΔMTQ  ...(AAS postulate)

The corresponding parts of the congruent triangles are congruent

∴ TL = TM   ...(C.P.C.T.)

Hence, T is equidistant from PQ and QR.