Advertisements
Advertisements
प्रश्न
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
Advertisements
उत्तर
Construction: From P, draw PL ⊥ AB and PM ⊥ BC
Proof: In ΔPLB and ΔPMB
∠PLB = ∠PMB ...(Each = 90°)
∠PBL = ∠PBM ...(Given)
PB = PB ...(Common)
∴ By Angle – angle side criterion of congruence,
ΔPLB ≅ ΔPMB ...(AAS postulate)
The corresponding parts of the congruent triangles are congruent
∴ PL = PM ...(C.P.C.T.)
Hence, P is equidistant from AB and BC
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.
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 points at a distance 2 cm from a fixed line.
Describe the locus of the moving end of the minute hand of a clock.
Describe the locus of points inside a circle and equidistant from two fixed points on the circumference of the circle.
Describe the locus of points at distances less than or equal to 2.5 cm from a given point.
In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.
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.
Given: ∠BAC, a line intersects the arms of ∠BAC in P and Q. How will you locate a point on line segment PQ, which is equidistant from AB and AC? Does such a point always exist?
