Question

In the adjoining figure, $segXY||segAC$, if $3AX=2BX$ and $XY=9$ then find the length of $AC$

Answer 1

REF.Image

Given XY is parallel to AC i.e, $XY\parallel AC$

and BC traverses through XY and AC, thus the corresponding angles are equal

i.e, $\angle XYB=\angle ACB...\left(i\right)$

similarly AB traverses through XY and AC.

$\therefore \angle BXY=\angle BAC...\left(ii\right)$

Now, In $\Delta ABC$ and $\Delta XBY$

$\angle B=\angle B$ [same angle]

$\angle ACB=\angle XYB$ [from (i)]

$\angle BAC=\angle BXY$ [from (ii)]

By $\Delta AA$ criterion of similarity, $\Delta ABC\sim \Delta XBY$

Since $\Delta ABC\sim \Delta XBY$

$\Rightarrow \frac{AB}{XB}=\frac{AC}{XY}$

$\Rightarrow AC=\frac{AB}{XB}.XY=\frac{(XB+XA).XY}{XB}$

It is given that $3AX=2BX$ and $XY=9$

$\Rightarrow AX=\frac{2}{3}BX$

$\therefore AC=\frac{(BX+\frac{2}{3}BX).9}{BX}=(1+\frac{2}{3}).9=\frac{(3+2)}{3}.9$

$\Rightarrow AC=15$