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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

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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.
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संबंधित प्रश्न
Describe the locus of vertices of all isosceles triangles having a common base.
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 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.
In Δ PQR, s is a point on PR such that ∠ PQS = ∠ RQS . Prove thats is equidistant from PQ and QR.
In Δ ABC, B and Care fixed points. Find the locus of point A which moves such that the area of Δ ABC remains the same.
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The locus of points inside a circle and equidistant from two fixed points on the circle.
Using only a ruler and compass construct ∠ABC = 120°, where AB = BC = 5 cm.
(i) Mark two points D and E which satisfy the condition that they are equidistant from both ABA and BC.
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- Mark the point of intersection of the loci with the letter P and measure PC.
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Use ruler and compasses only for the following questions:
Construct triangle BCP, when CB = 5 cm, BP = 4 cm, ∠PBC = 45°.
Complete the rectangle ABCD such that :
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