Pria wrote:

Explain Please

Here given BD is parallel to AE,

so angle CBD = angle CAE, and angle BDC = angle AEC.

Therefore the triangles BCD and ACE are similar (Angle C is common to both triangles and by AAA triangle BCD and triangle ACE are similar)

Now it is given side BC = x , AB = y and AC =x+w.

side CD = y, DE = z and CE = y+Z

Now as both triangles BCD and ACE are similar

therfore we have

\(\frac{CB}{CA}\) = \(\frac{CD}{CE}\)

substitute the values we get

\(\frac{x}{(x+w)}\)= \(\frac{y}{(y+z)}\)

or x(y+z) = y(x+w)

or xy + xz = xy + wy

or xz = wy. So option C.

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