Mathematics
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.
Answer
Steps of construction:
Draw a parallelogram such that AB > BC.
Join AC the diagonal of parallelogram.
Draw BX, the angle bisector of ∠ABC and let it intersect AC at P.
From P, draw PL ⊥ AB and PM ⊥ BC.
In ∆PLB and ∆PMB,
⇒ ∠PLB = ∠PMB [Each 90°]
⇒ ∠PBL = ∠PBM [As BX is angle bisector of ABC.]
⇒ PB = PB [Common]
∴ ∆PLB ≅ ∆PMB by AAS axiom.
∴ PL = PM
Hence, proved that P is equidistant from AB and BC.
Related Questions
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that:
(i) point A is equidistant from all the three sides of the triangle.
(ii) AM bisects angle LMN.
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.
Construct a triangle ABC in which angle ABC = 75°, AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P; prove that P is equidistant from B and C; and also from AC and BC.
Use ruler and compasses only for this question.
(i) Construct △ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.