pranab01 wrote:
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.
If you put the reason, it will be better---
If two triangles are similar, the ratio of their corresponding sides are equal.
---------------------------------------------------------------------------------------------------------
Please let me know, if I am wrong.