Advertisements
Advertisements
प्रश्न
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 AB and AC.
Advertisements
उत्तर
Construction: Draw LF ⊥ AC
Proof: In ΔAFL and ΔAFE,
∠FEA = ∠FLA ...(Each = 90°)
∠LAF = FAE ...(AD bisects ∠BAC)
AF = AF ...(Common)
∴ By angle – Angle side criterion of congruence,
ΔAFL ≅ AFE ...(AAS postulate)
The corresponding parts of the congruent triangles are congruent.
∴ FE = FL ...(C.P.C.T.)
Hence, F is equidistant from AB and AC.
संबंधित प्रश्न
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 the centre of a wheel of a bicycle going straight along a level road.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Describe the locus of the door handle, as the door opens.
Describe the locus of points at distances greater than or equal to 35 mm from a given point.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
Find the locus of points which are equidistant from three non-collinear points.
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.
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.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect in P. Show that P is equidistant from the opposite sides AB and CD.
