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Question Number 81887 by lalitchand last updated on 16/Feb/20

Commented by mr W last updated on 16/Feb/20

$$let\frac{FA}{BA}=k,\frac{1}{2}<k<1$$$$DF=\frac{BA}{2}$$$$\frac{OD}{OF}=\frac{DF}{FA}=\frac{BA}{2\times k\times BA}=\frac{1}{2k}$$$$i.e.\frac{OD}{OD+DF}=\frac{1}{2k}$$$$\Rightarrow (2k-1)od=df$$$$\Rightarrow \frac{OD}{DF}=\frac{1}{2k-1}$$$$\frac{\mathrm{\Delta }OCD}{\mathrm{\Delta }BFD}=\frac{OD}{DF}=\frac{1}{2k-1}\ne 2always$$$$\frac{1}{2k-1}=2\Rightarrow k=\frac{3}{4}$$$$i.e.onlywhen\frac{BF}{FB}=\frac{1}{3},2\mathrm{\Delta }BFD=\mathrm{\Delta }OCD.$$

Commented by mr W last updated on 16/Feb/20

$${it}^{\prime }snotgenerallytrue!$$$$pleasecheckthequestionagain!$$

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