In the given figure,BC = AC,angle AFD = 40^{c | vedclass

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(D) From $	riangle AFD$,we have $\angle A + \angle AFD + \angle D = 180^{\circ}$.Let $\angle A = \alpha$ and $\angle D = \beta$. Then $\alpha + 40^{\

Vedclass · Accepted Answer

$100$

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(D) From $	riangle AFD$,we have $\angle A + \angle AFD + \angle D = 180^{\circ}$.Let $\angle A = \alpha$ and $\angle D = \beta$. Then $\alpha + 40^{\circ} + \beta = 180^{\circ}$,so $\alpha + \beta = 140^{\circ}$.In $	riangle ABC$,since $BC = AC$,we have $\angle ABC = \angle BAC = \alpha$. Thus,$\angle ACB = 180^{\circ} - 2\alpha$.In $	riangle CED$,since $CE = CD$,we have $\angle CED = \angle CDE = \beta$. Thus,$\angle ECD = 180^{\circ} - 2\beta$.Since $A, C, D$ lie on a straight line,$\angle ACB + \angle BCE + \angle ECD = 180^{\circ}$.Substituting the values: $(180^{\circ} - 2\alpha) + \angle BCE + (180^{\circ} - 2\beta) = 180^{\circ}$.$\angle BCE = 2\alpha + 2\beta - 180^{\circ} = 2(\alpha + \beta) - 180^{\circ}$.Since $\alpha + \beta = 140^{\circ}$,$\angle BCE = 2(140^{\circ}) - 180^{\circ} = 280^{\circ} - 180^{\circ} = 100^{\circ}$.

In the given figure,$BC = AC$,$\angle AFD = 40^{\circ}$ and $CE = CD$. Then the value of $\angle BCE$ is equal to ......$^{\circ}$

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