Given: XY ∥ AC

ar (∆ XBY) = ar (XACY) ……….(i)

To show:

Consider ∆ ABC, XY ∥ AC

So, Using Basic Proportionality theorem, we have

……….(ii)

Now, in ∆ XBY and ∆ ABC,

∠XBY = ∠ABC [common angle]

[Using (ii)]

⇒ ∆ XBY ∼ ∆ ABC [By SAS criterion]

Now, we know that the ratios of the areas of two similar triangles are equal to the ratio of squares of any two corresponding sides.

From (i), we have

ar (∆ XBY) = ar (XACY)

Let ar (∆ XBY) = x = ar (XACY) ⇒ ar (∆ ABC) = ar (∆ XBY) + ar (XACY) = x + x = 2x

Now, we know that

XB = AB – AX

Rationalizing the denominator, we have