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
$${let}\:{AC}={p},\:{BC}={q},\:{AB}={s}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$$${let}\:{AD}={u},\:{DE}={w},\:{EB}={v} \\ $$$${due}\:{to}\:{similarity}\:{of}\:{the}\:{triangles}, \\ $$$$\frac{{a}}{{u}}=\frac{{b}}{{v}}=\frac{{c}}{{w}}=\frac{{r}}{{s}} \\ $$$$\Delta=\frac{{pq}}{\mathrm{2}}=\frac{\left({p}+{q}+{s}\right){r}}{\mathrm{2}}\:\Rightarrow{r}=\frac{{pq}}{{p}+{q}+{s}} \\ $$$$\frac{{a}}{{r}}=\frac{{u}}{{s}}=\frac{{p}−\mathrm{2}{r}}{{p}}=\frac{{p}−{q}+{s}}{{p}+{q}+{s}} \\ $$$$\frac{{b}}{{r}}=\frac{{v}}{{s}}=\frac{{q}−\mathrm{2}{r}}{{q}}=\frac{−{p}+{q}+{s}}{{p}+{q}+{s}} \\ $$$$\frac{{c}}{{r}}=\frac{{w}}{{s}}=\frac{{s}−{u}−{v}}{{s}}=\frac{{p}+{q}−{s}}{{p}+{q}+{s}} \\ $$$$\frac{{a}}{{r}}+\frac{{b}}{{r}}+\frac{{c}}{{r}}=\frac{{u}}{{s}}+\frac{{v}}{{s}}+\frac{{w}}{{s}}=\mathrm{1} \\ $$$$\Rightarrow{a}+{b}+{c}={r} \\ $$$$\frac{{ab}}{{r}^{\mathrm{2}} }=\frac{{p}−{q}+{s}}{{p}+{q}+{s}}×\frac{−{p}+{q}+{s}}{{p}+{q}+{s}}=\frac{{s}^{\mathrm{2}} −\left({p}−{q}\right)^{\mathrm{2}} }{\left({p}+{q}+{s}\right)^{\mathrm{2}} }=\frac{\mathrm{2}{pq}}{\left({p}+{q}+{s}\right)^{\mathrm{2}} } \\ $$$$\frac{{c}}{{r}}=\frac{\left({p}+{q}−{s}\right)\left({p}+{q}+{s}\right)}{\left({p}+{q}+{s}\right)^{\mathrm{2}} }=\frac{\left({p}+{q}\right)^{\mathrm{2}} −{s}^{\mathrm{2}} }{\left({p}+{q}+{s}\right)^{\mathrm{2}} }=\frac{\mathrm{2}{pq}}{\left({p}+{q}+{r}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{ab}}{{r}^{\mathrm{2}} }=\frac{{c}}{{r}}\:\Rightarrow{ab}={cr} \\ $$$$\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ca}}=\frac{{a}+{b}+{c}}{{abc}}=\frac{{r}}{{c}^{\mathrm{2}} {r}}=\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 21/Nov/22
$${thank}\:{you}\:{sir}\:{W} \\ $$$${grateful} \\ $$