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Sum of Opposite Angles of a Cyclic Quadrilateral

In the given ...

Question

In the given figure, $∠$ BAD = $65^{o}$ , $∠$ ABD = $70^{o}$ and $∠$ BDC = $45^{o}$ , Find :

(i) $∠$ BCD

(ii) $∠$ ACB

Hence, show that AC is a diameter.

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Solution

(i) $\angle BCD = 180^o-\angle BAD = 180^o-65^o=115^o$ ( $ABCD$ is a cyclic quadrilateral)

(ii) In $\triangle ABD$

$\angle ADB = 180^o-65^o-70^o=45^o$

$\angle ACB = \angle ADB = 45^o$ (angles in the same segment)

$\angle ADC = \angle ADB + \angle BDC = 45^o+45^o = 90^o$

Therefore $AC$ is the diameter since the angle subtended by $AC$ on the circumference is $90^o$ .

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