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in Mathematics by (54.1k points)

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.

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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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