Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर
A is equidistant from B and 0 . Therefore, A lies on perpendicular bisector of BO.
C is equidistant from Band 0. Therefore, C lies on perpendicular bisector of BO.
A and C both lie on perpendicular bisector of BO.
Hence, AC is perpendicular bisector of BO.
Since AC is perpendicular bisector of BO so ∠ AMB = ∠ AMO = right angle.
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.
On a graph paper, draw the lines x = 3 and y = –5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Plot the points A(2, 9), B(–1, 3) and C(6, 3) on graph paper. On the same graph paper draw the locus of point A so that the area of ΔABC remains the same as A moves.
AB and CD are two intersecting lines. Find a point equidistant from AB and CD, and also at a distance of 1.8 cm from another given line EF.
In Δ PQR, s is a point on PR such that ∠ PQS = ∠ RQS . Prove thats is equidistant from PQ and QR.
In the given figure ABC is a triangle. CP bisects angle ACB and MN is perpendicular bisector of BC. MN cuts CP at Q. Prove Q is equidistant from B and C, and also that Q is equidistant from BC and AC.
A and B are fixed points while Pis a moving point, moving in a way that it is always equidistant from A and B. What is the locus of the path traced out by the pcint P?
Without using set squares or a protractor, construct:
- Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
- Draw the locus of a point which moves so that it is always 2.5 cm from B.
- Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
- Mark the point of intersection of the loci with the letter P and measure PC.
Use ruler and compass to answer this question. Construct ∠ABC = 90°, where AB = 6 cm, BC = 8 cm.
- Construct the locus of points equidistant from B and C.
- Construct the locus of points equidistant from A and B.
- Mark the point which satisfies both the conditions (a) and (b) as 0. Construct the locus of points keeping a fixed distance OA from the fixed point 0.
- Construct the locus of points which are equidistant from BA and BC.
