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 A and B.
Advertisements
उत्तर
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
संबंधित प्रश्न
Construct a right angled triangle PQR, in which ∠Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
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.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Draw an ∠ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC; and also equidistant from A and B.
In the figure given below, find a point P on CD equidistant from points A and B.
Describe the locus of points at a distance 2 cm from a fixed line.
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.
