# In the given figu

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

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