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Question Number 181023 by mnjuly1970 last updated on 20/Nov/22

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

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

$$letAC=p,BC=q,AB=s=\sqrt{p^2+q^2}$$$$letAD=u,DE=w,EB=v$$$$duetosimilarityofthetriangles,$$$$\frac{a}{u}=\frac{b}{v}=\frac{c}{w}=\frac{r}{s}$$$$\mathrm{\Delta }=\frac{pq}{2}=\frac{(p+q+s)r}{2}\Rightarrow r=\frac{pq}{p+q+s}$$$$\frac{a}{r}=\frac{u}{s}=\frac{p-2r}{p}=\frac{p-q+s}{p+q+s}$$$$\frac{b}{r}=\frac{v}{s}=\frac{q-2r}{q}=\frac{-p+q+s}{p+q+s}$$$$\frac{c}{r}=\frac{w}{s}=\frac{s-u-v}{s}=\frac{p+q-s}{p+q+s}$$$$\frac{a}{r}+\frac{b}{r}+\frac{c}{r}=\frac{u}{s}+\frac{v}{s}+\frac{w}{s}=1$$$$\Rightarrow a+b+c=r$$$$\frac{ab}{r^2}=\frac{p-q+s}{p+q+s}\times \frac{-p+q+s}{p+q+s}=\frac{s^2-{(p-q)}^2}{{(p+q+s)}^2}=\frac{2pq}{{(p+q+s)}^2}$$$$\frac{c}{r}=\frac{(p+q-s)(p+q+s)}{{(p+q+s)}^2}=\frac{{(p+q)}^2-s^2}{{(p+q+s)}^2}=\frac{2pq}{{(p+q+r)}^2}$$$$\Rightarrow \frac{ab}{r^2}=\frac{c}{r}\Rightarrow ab=cr$$$$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=\frac{r}{c^{2}r}=\frac{1}{c^2}✓$$

Commented by mnjuly1970 last updated on 21/Nov/22

$$thankyousirW$$$$grateful$$

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