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प्रश्न
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
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उत्तर
Given: Two interesting circles with centres C &D.
AB is their common chord.
To prove: AB bisected by CD at right angles.
Proof : CA = CB ...(radii)
∴ C lies on the right bisector of AB.
Similarly, D lies on the right bisector of AB.
Therefore, CD is the right bisector of AB.
Hence proved.
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.
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
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 moving end of the minute hand of a clock.
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 3 cm from a given point.
In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
ΔPBC, ΔQBC and ΔRBC are three isosceles triangles on the same base BC. Show that P, Q and R are collinear.
