Question

Answer the following questions (Any two)​

Answer 1

Answer:

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18. Join BC.

Then angle ACB = 90° [Angle in a semi circle]

Since DCBE is a cyclic quad.

angle BCD + angle BED = 180°

Adding angle ACB on both sides

angle BCD + angle BED +angle ACB = angle ACB + 180°

( angle BCD + angle ACB) + angle BED = 90° + 180°

angle ACD + angle BED = 270°

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23. Join BC , meeting AD at M.

In ΔDAM and ΔCAM,

AB = AC and ∠RAM = ∠CAM [given]

and AM = AM [common sides]

ΔBAM ≅ ΔCAM [by SAS congruency]

∴ BM = CM

and ∠BMA = ∠CMA [by CPCT] ---- i)

But ∠BMA + ∠CMA = 180° ----- (ii)

From Eqs. (i) and (ii), we get

∠BMA = ∠CMA = 90°

⇒ AD is the perpendicular bisector of chord BC, but the perpendicular bisector of a chord always passes through the centre of circle.

∴ AD passes through the centre O of circle.

O lies on AD. Hence proved.

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I have answered Q-: 18 & 23

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