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
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
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.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
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 points inside a circle and equidistant from two fixed points on the circumference of the circle.
Describe the locus of the centres of all circles passing through two fixed points.
The speed of sound is 332 metres per second. A gun is fired. Describe the locus of all the people on the earth’s surface, who hear the sound exactly one second later.
Describe the locus of points at distances greater than 4 cm from a given point.
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
ΔPBC, ΔQBC and ΔRBC are three isosceles triangles on the same base BC. Show that P, Q and R are collinear.
