Advertisements
Advertisements
प्रश्न
In given figure, ABCD is a kite. AB = AD and BC =CD. Prove that the diagona AC is the perpendirular bisector of the diagonal BD.
Advertisements
उत्तर
A is equidistant from B and D. Therefore, A lies on perpendicular bisector of BD.
C is equidistant from Band D. Therefore, C lies on perpendicular bisector ofBD.
A and C both lie on perpendicular bisector of BD.
Hence, AC is perpendicular bi sector of BD.
APPEARS IN
संबंधित प्रश्न
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°
Hence:
1) Construct the locus of points equidistant from BA and BC
2) Construct the locus of points equidistant from B and C.
3) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Describe the locus of a point P, so that:
AB2 = AP2 + BP2,
where A and B are two fixed points.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
- Complete the rectangle ABCD such that:
- P is equidistant from AB and BC.
- P is equidistant from C and D.
- Measure and record the length of AB.
Construct a rhombus ABCD whose diagonals AC and BD are 8 cm and 6 cm respectively. Find by construction a point P equidistant from AB and AD and also from C and D.
In given figure 1 ABCD is an arrowhead. AB = AD and BC = CD. Prove th at AC produced bisects BD at right angles at the point M
Draw and describe the locus in the following case:
The locus of a point in the rhombus ABCD which is equidistant from the point A and C.
Using a ruler and compass only:
(i) Construct a triangle ABC with BC = 6 cm, ∠ABC = 120° and AB = 3.5 cm.
(ii) In the above figure, draw a circle with BC as diameter. Find a point 'P' on the circumference of the circle which is equidistant from Ab and BC.
Measure ∠BCP.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
