Q. 184.5( 2 Votes )

In the given figu

Answer :

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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