Q5 of 23 Page 270

In the adjacent figure, AB is a chord of circle with centre O. CD is the diameter perpendicualr to AB. Show that AD = BD.


Given, AB is a chord of circle with centre O. CD is the diameter


perpendicular to AB.


We know that line drawn from the center of a circle to the chord


Perpendicular to it bisects the chord.


AP = BP


In ADP and BDP,


AP = BP


APD = BPD = 90


PD = PD (Common)


ADP BDP


AD = BD (CPCT)


More from this chapter

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3

If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

 4

If two intersecting chords of a circle make equal angles with diameter passing through their point of intersection, prove that the chords are equal.

 1

In the figure, ‘O’ is the centre of the circle. ∠AOB = 100° find ∠ADB.

 2

In the figure, ∠BAD = 40° then find ∠BCD.