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Angles in Same Segment of a Circle

In the figure...

Question

In the figure, given below, AD = BC, $∠ B A C = 30^{\circ} a n d ∠ C B D = 70^{\circ} .$

Find :

(i) $∠ B C D$

(ii) $∠ B C A$

(iii) $∠ A B C$

(iv) $∠ A D B$

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Solution

From $chord \ CD: \angle DBC = \angle DAC$ ( angles in the same segment)

$\Rightarrow \angle DAC = 70^o$

From $chord \ CB: \angle CAB = \angle CDB$ (angles in the same segment)

$\Rightarrow \angle CDB = 30^o$

In $\triangle \angle DCB: 70 + 30 + \angle DCB = 180 \Rightarrow \angle DCB = 80^o$

Since $AD = BC: \angle DCA = \angle CAB \Rightarrow \angle DCA = 30^o$

In $\triangle ADC: 70 + \angle ADB + 30 + 30 = 180 \Rightarrow \angle ADB = 50^o$

$ABCD$ is a cyclic quadrilateral. $\therefore 50 + 30 + \angle DBA + 70 = 180 \Rightarrow DBA = 30^o$

Hence

(i) $\angle BCD = 80^o$

(ii) $\angle BCA = 50^o$

(iii) $\angle ABC = 100^o$

(iv) $\angle ADB = 50^o$

$\\$

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