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प्रश्न
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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उत्तर
Steps of Construction:
- Draw a line segment QR = 4.5 cm
- At Q, draw a ray QX making an angle of 90°
- With centre R and radius 8 cm, draw an arc which intersects QX at P.
- Join RP. ΔPQR is the required triangle.
- Draw the bisector of ∠PQR that meets PR in T.
- 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.
संबंधित प्रश्न
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.
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