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
Steps of Construction:
i) Draw a line segment QR = 4.5 cm
ii) At Q, draw a ray QX making an angle of 90°
iii) With centre R and radius 8 cm, draw an arc which intersects QX at P.
iv) Join RP.
ΔPQR 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 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 (CPCT)
Hence, T is equidistant from PQ and QR.
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