Question

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

Answer 1

We have, $BD=DE=EC=\frac{1}{3}a$

$\ln △ABE,\frac{\sin (x+y)}{\frac{2}{3}a}=\frac{\sin B}{AE}$ ...(1)

$\ln △ADC,\frac{\sin (y+z)}{\frac{2}{3}a}=\frac{\sin C}{AD}$ ...(2)

$\ln △ABD,\frac{\sin x}{\frac{1}{3}a}=\frac{\sin B}{AD}$ ...(3)

$\ln △AFC,\frac{\sin z}{\frac{1}{3}a}=\frac{\sin C}{AE}$ ...(4)

From (1), (2), (3), and (4), we have

$\frac{\sin (x+y)\sin (y+z)}{\sin x\sin z}=4.$