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 A and B.
Advertisements
Solution
Construction: Join FB and FC
Proof: In ΔAFE and ΔFBE,
AE = EB ...(E is the mid-point of AB)
∠FEA = ∠FEB ...(Each = 90°)
FE = FE ...(Common)
∴ By side Angle side criterion of congruence,
ΔAFE ≅ ΔFBE ...(SAS Postulate)
The corresponding parts of the congruent triangles are congruent.
∴ AF = FB ...(C.P.C.T.)
Hence, F is equidistant from A and B.
RELATED QUESTIONS
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that:
- Point A is equidistant from all the three sides of the triangle.
- AM bisects angle LMN.
Describe the locus of a stone dropped from the top of a tower.
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 points at distances less than 3 cm from a given point.
Describe the locus of points at distances greater than 4 cm from a given point.
Describe the locus of points at distances less than or equal to 2.5 cm from a given point.
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 .
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
ΔPBC and ΔQBC are two isosceles triangles on the same base. Show that the line PQ is bisector of BC and is perpendicular to BC.
