77
In the given figure, Z is a point on the side BC of $\triangle ABC$ such that XZ $||$ AB and YZ $||$ AC. If XY and CB produced meet at O, then prove that $ZO^2 = OB \times OC$.
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In $\triangle OZX$, ZX $||$ BY (As XZ $||$ AB)
$\therefore \frac{OB}{OZ} = \frac{OY}{OX}$ --- (1)
In $\triangle OCX$, ZY $||$ CX (As YZ $||$ AC)
$\therefore \frac{OZ}{OC} = \frac{OY}{OX}$ --- (2)
Using (1) and (2), we get
$\frac{OB}{OZ} = \frac{OZ}{OC}$
$\Rightarrow OZ^2 = OB \times OC$
$\therefore \frac{OB}{OZ} = \frac{OY}{OX}$ --- (1)
In $\triangle OCX$, ZY $||$ CX (As YZ $||$ AC)
$\therefore \frac{OZ}{OC} = \frac{OY}{OX}$ --- (2)
Using (1) and (2), we get
$\frac{OB}{OZ} = \frac{OZ}{OC}$
$\Rightarrow OZ^2 = OB \times OC$