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प्रश्न
ΔPBC and ΔQBC are two isosceles triangles on the same base. Show that the line PQ is bisector of BC and is perpendicular to BC.
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उत्तर
Given: ΔPBC and ΔQBC are two isosceles triangles on the same base BC.
To prove: Line PQ is the perpendicular bisector of BC.
Proof: In ΔPBC, PB = PC
Since, the locus of a point equidistant from B and C is the perpendicular bisector of 1 of the line segment BC
∴ P lies on 1
Similarly Q lies on 1
Therefore, PQ is the perpendicular bisector of BC.
Hence proved.
संबंधित प्रश्न
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.
Use ruler and compasses only for this question.
- Construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
- Construct the locus of points inside the triangle which are equidistant from BA and BC.
- Construct the locus of points inside the triangle which are equidistant from B and C.
- Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
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.
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 an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
In the figure given below, find a point P on CD equidistant from points 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.
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.
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.
