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प्रश्न
Describe the locus of points inside a circle and equidistant from two fixed points on the circumference of the circle.
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उत्तर
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.
संबंधित प्रश्न
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.
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 ∠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.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
- AB and BC;
- B and D.
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.
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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.
