Advertisements
Advertisements
Question
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
Solution
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
APPEARS IN
RELATED QUESTIONS
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.
Prove that:
F is equidistant from A and B.
Draw an ∠ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC; and also equidistant from A and B.
In the figure given below, find a point P on CD equidistant from points A and B.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that:
- P is equidistant from B and C.
- P is equidistant from AB and BC.
- Measure and record the length of PB.
Describe the locus for questions 1 to 13 given below:
1. The locus of a point at a distant 3 cm from a fixed point.
Describe the locus of a stone dropped from the top of a tower.
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
ΔPBC and ΔQBC are two isosceles triangles on the same base BC but on the opposite sides of line BC. Show that PQ bisects BC at right angles.
ΔPBC, ΔQBC and ΔRBC are three isosceles triangles on the same base BC. Show that P, Q and R are collinear.
