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Question Number 206053 by mr W last updated on 05/Apr/24

Commented by mr W last updated on 05/Apr/24

$$anunsolvedoldquestion\left(Q182165\right)$$

Answered by mr W last updated on 05/Apr/24

Commented by mr W last updated on 06/Apr/24

$$\frac{AE}{BE}=\frac{CF}{DF}=\frac{1}{k},say$$$$\frac{\mathrm{\Delta }BCF}{\mathrm{\Delta }BDF}=\frac{CF}{DF}=\frac{1}{k}$$$$\Rightarrow k(S_1+S_3)=\mathrm{\Delta }DBF...\left(i\right)$$$$\frac{\mathrm{\Delta }DAE}{\mathrm{\Delta }DBE}=\frac{AE}{BE}=\frac{1}{k}$$$$\Rightarrow k(S_2+S_6)=\mathrm{\Delta }DBE...\left(ii\right)$$$$\left(i\right)+\left(ii\right):$$$$k(S_1+S_3+S_2+S_6)=\mathrm{\Delta }DBF+\mathrm{\Delta }DBE$$$$\Rightarrow k(S_1+S_3+S_2+S_6)=S+S_4+S_5...\left(iii\right)$$$$similarly$$$$\Rightarrow S_1+S_5+S_2+S_4=k(S+S_3+S_6)...\left(iv\right)$$$$\left(iii\right)+\left(iv\right):$$$$k(S_1+S_3+S_2+S_6)+S_1+S_5+S_2+S_4=S+S_4+S_5+k(S+S_3+S_6)$$$$(k+1)(S_1+S_2)=(k+1)S$$$$\Rightarrow S=S_1+S_2✓$$

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