Points D and E are taken on the side BC of a | vedclass

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(C) Let $BD = DE = EC = k$. Then $BE = 2k$,$DC = 2k$,$BC = 3k$.Using the sine rule in $\Delta ABD$,$\frac{\sin x}{BD} = \frac{\sin B}{AD} \implies \fr

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$4$

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(C) Let $BD = DE = EC = k$. Then $BE = 2k$,$DC = 2k$,$BC = 3k$.Using the sine rule in $\Delta ABD$,$\frac{\sin x}{BD} = \frac{\sin B}{AD} \implies \frac{\sin x}{k} = \frac{\sin B}{AD} \implies \frac{AD}{\sin B} = \frac{k}{\sin x}$.In $\Delta ABE$,$\frac{\sin(x + y)}{BE} = \frac{\sin B}{AE} \implies \frac{\sin(x + y)}{2k} = \frac{\sin B}{AE} \implies \frac{AE}{\sin B} = \frac{2k}{\sin(x + y)}$.In $\Delta AEC$,$\frac{\sin z}{EC} = \frac{\sin C}{AE} \implies \frac{\sin z}{k} = \frac{\sin C}{AE} \implies \frac{AE}{\sin C} = \frac{k}{\sin z}$.In $\Delta ADC$,$\frac{\sin(y + z)}{DC} = \frac{\sin C}{AD} \implies \frac{\sin(y + z)}{2k} = \frac{\sin C}{AD} \implies \frac{AD}{\sin C} = \frac{2k}{\sin(y + z)}$.Now,consider the ratio $\frac{\sin(x + y)\sin(y + z)}{\sin x \sin z} = \left( \frac{\sin(x + y)}{\sin x} ight) \left( \frac{\sin(y + z)}{\sin z} ight)$.From the sine rule ratios: $\frac{\sin(x + y)}{\sin x} = \frac{2k/AE}{k/AD} = \frac{2AD}{AE}$ and $\frac{\sin(y + z)}{\sin z} = \frac{2k/AD}{k/AE} = \frac{2AE}{AD}$.Multiplying these: $\frac{2AD}{AE} 	imes \frac{2AE}{AD} = 4$.

Points $D$ and $E$ are taken on the side $BC$ of a triangle $ABC$ such that $BD = DE = EC$. If $\angle BAD = x$,$\angle DAE = y$,and $\angle EAC = z$,then the value of $\frac{\sin(x + y)\sin(y + z)}{\sin x \sin z}$ is:

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