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.
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Please let me know, if I am wrong.