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 AB and AC.
Advertisements
उत्तर
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.
संबंधित प्रश्न
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that:
- P is equidistant from B and C.
- P is equidistant from AB and BC.
- Measure and record the length of PB.
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.
A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
- always 4 cm from the line AB.
- equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
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 Fig. ABCD is a quadrilateral in which AB = BC. E is the point of intersection of the right bisectors of AD and CD. Prove that BE bisects ∠ABC.
Find the locus of points which are equidistant from three non-collinear points.
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.
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.
Use ruler and compasses for the following question taking a scale of 10 m = 1 cm. A park in a city is bounded by straight fences AB, BC, CD and DA. Given that AB = 50 m, BC = 63 m, ∠ABC = 75°. D is a point equidistant from the fences AB and BC. If ∠BAD = 90°, construct the outline of the park ABCD. Also locate a point P on the line BD for the flag post which is equidistant from the corners of the park A and B.
