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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 ((FA)/(BA))=k, (1/2)<k<1 DF=((BA)/2) ((OD)/(OF))=((DF)/(FA))=((BA)/(2×k×BA))=(1/(2k)) i.e. ((OD)/(OD+DF))=(1/(2k)) ⇒(2k−1)od=df ⇒((OD)/(DF))=(1/(2k−1)) ((ΔOCD)/(ΔBFD))=((OD)/(DF))=(1/(2k−1))≠2 always (1/(2k−1))=2 ⇒k=(3/4) i.e. only when ((BF)/(FB))=(1/3), 2ΔBFD=ΔOCD.

$${let}\:\frac{{FA}}{{BA}}={k},\:\frac{\mathrm{1}}{\mathrm{2}}<{k}<\mathrm{1} \\ $$$${DF}=\frac{{BA}}{\mathrm{2}} \\ $$$$\frac{{OD}}{{OF}}=\frac{{DF}}{{FA}}=\frac{{BA}}{\mathrm{2}×{k}×{BA}}=\frac{\mathrm{1}}{\mathrm{2}{k}} \\ $$$${i}.{e}.\:\frac{{OD}}{{OD}+{DF}}=\frac{\mathrm{1}}{\mathrm{2}{k}} \\ $$$$\Rightarrow\left(\mathrm{2}{k}−\mathrm{1}\right){od}={df} \\ $$$$\Rightarrow\frac{{OD}}{{DF}}=\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}} \\ $$$$\frac{\Delta{OCD}}{\Delta{BFD}}=\frac{{OD}}{{DF}}=\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\neq\mathrm{2}\:{always} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}=\mathrm{2}\:\Rightarrow{k}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${i}.{e}.\:{only}\:{when}\:\frac{{BF}}{{FB}}=\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{2}\Delta{BFD}=\Delta{OCD}. \\ $$

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

it′s not generally true! please check the question again!

$${it}'{s}\:{not}\:{generally}\:\:{true}! \\ $$$${please}\:{check}\:{the}\:{question}\:{again}! \\ $$

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