Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Given: Two ΔSPBC and QBC on the same base BC but in the opposite sides of BC such that PB = PC and QB = QC.
To prove: PQ bisects BC and is ⊥ to BC.
Proof: Since, the locus of points equidistant from two given points is the perpendicular bisector of the segment joining them. Therefore, ΔPBC is isoceles
⇒ P lies on the perpendicular bisector of BC
ΔQBC is isoceles ⇒ QB = QC
⇒ Q lies on the perpendicular bisectors of BC
∴ PQ is the perpendicular bisectors of BC
Hence, PQ bisects BC at right angles.
Hence proved.
APPEARS IN
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.
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 given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
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 greater than or equal to 35 mm from a given point.
In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.
In a quadrilateral PQRS, if the bisectors of ∠ SPQ and ∠ PQR meet at O, prove that O is equidistant from PS and QR.
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
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.
