Q9 of 28 Page 107

In the given figure ∠ABC is given. Find the locus of points equidistant from BA and BC and lying in the interior part of ∠ABC.



Draw angle bisector BX of ∠ABC.


Take any point P on BX. Now draw ⊥ on AB and BC.


PL ⊥ AB and PM ⊥ BC


∴ ∠PLB = ∠PMB = 90°


In Δ PLB and Δ PMB,


∠PLB = ∠PMB (by construction)


∠LBP = ∠PBM (BP is bisector of ∠B)


BP = BP (common)


∴ Δ PLB ≅ Δ PMB (By AAS congruency)


⇒ PL = PM (By CPCT)


Thus, point P is the interior of ∠ABC and equidistant from AB and BC. So P is the locus.


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Find the locus of point ℓ equidistant from three vertices and three sides of a triangle.

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In a ΔABC the median AD, BE and CF intersect at a point G. If AG = 6 cm, BE = 9 cm and GF = 4.5 cm, then find GD, BG and CF.