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Question Number 170148 by Shrinava last updated on 17/May/22

Answered by mr W last updated on 20/May/22

Commented by mr W last updated on 20/May/22

$${it}^{\prime }seasytoseethatallhatched$$$$trianglesaresimilar.$$$$saythesidelengthofboth$$$$equilateralsiss.$$$$q+e+\frac{c}{a}p=s...\left(1\right)$$$$\frac{c}{a}q+f+\frac{b}{a}p=s...\left(2\right)$$$$\frac{b}{a}q+d+p=s...\left(3\right)$$$$\mathrm{\Sigma }:$$$$(d+e+f)+(a+b+c)\left(\frac{p+q}{a}\right)=3s...\left(i\right)$$$$\frac{d}{a}p+a+\frac{e}{a}q=s...\left(4\right)$$$$\frac{e}{a}p+c+\frac{f}{a}q=s...\left(5\right)$$$$\frac{f}{a}p+b+\frac{d}{a}q=s...\left(6\right)$$$$\mathrm{\Sigma }:$$$$(a+b+c)+(d+e+f)\left(\frac{p+q}{a}\right)=3s...\left(ii\right)$$$$\left(i\right)-\left(ii\right):$$$$(d+e+d)(1-\frac{p+q}{a})-(a+b+c)(1-\frac{p+q}{a})=0$$$$(d+e+d)(1-\frac{p+q}{a})=(a+b+c)(1-\frac{p+q}{a})$$$$p+q>a\Rightarrow 1-\frac{p+q}{a}\ne 0$$$$\Rightarrow \mathit{d}+\mathit{e}+\mathit{f}=\mathit{a}+\mathit{b}+\mathit{c}\Rightarrow proved✓$$

Commented by Shrinava last updated on 21/May/22

$$\mathrm{Fantastic}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{yo}\mathrm{so}\mathrm{much}$$

Commented by Tawa11 last updated on 08/Oct/22

$$\mathrm{Great}\mathrm{sir}.$$

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