Advertisements
Advertisements
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.
Advertisements
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
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
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.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Describe the locus of the centres of all circles passing through two fixed points.
In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.
In a quadrilateral PQRS, if the bisectors of ∠ SPQ and ∠ PQR meet at O, prove that O is equidistant from PS and QR.
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
In Fig. ABCD is a quadrilateral in which AB = BC. E is the point of intersection of the right bisectors of AD and CD. Prove that BE bisects ∠ABC.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
In Fig. AB = AC, BD and CE are the bisectors of ∠ABC and ∠ACB respectively such that BD and CE intersect each other at O. AO produced meets BC at F. Prove that AF is the right bisector of BC.
