Advertisements
Advertisements
Question
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
Solution
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.
APPEARS IN
RELATED QUESTIONS
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 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.
By actual drawing obtain the points equidistant from lines m and n; and 6 cm from a point P, where P is 2 cm above m, m is parallel to n and m is 6 cm above n.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
- the locus of the centres of all circles which touch AB and AC,
- the locus of the centres of all the circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC .
In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
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.
