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.
APPEARS IN
RELATED QUESTIONS
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
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.
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.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Describe the locus of the door handle, as the door opens.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
- AB and BC;
- B and D.
Describe the locus of points at distances less than or equal to 2.5 cm from a given point.
In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.
Show that the locus of the centres of all circles passing through two given points A and B, is the perpendicular bisector of the line segment AB.
ΔPBC, ΔQBC and ΔRBC are three isosceles triangles on the same base BC. Show that P, Q and R are collinear.
