Question Number 146000 by mnjuly1970 last updated on 10/Jul/21
Commented by mr W last updated on 10/Jul/21
$${i}\:{think}\:{the}\:{area}\:{of}\:\Delta{ABC}\:{can}'{t}\:{be}\: \\ $$$${uniquely}\:{determined}\:{with}\:{the}\:{given}\: \\ $$$${condition}.\:{please}\:{recheck}. \\ $$
Commented by mr W last updated on 10/Jul/21
$${AC}={a} \\ $$$${BC}={b} \\ $$$${AB}={c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${r}=\frac{{ab}}{{a}+{b}+{c}} \\ $$$${AD}={a}−{r} \\ $$$${BD}={b}−{r} \\ $$$${AD}×{BD}={DE}^{\mathrm{2}} ={h}^{\mathrm{2}} \:\:\:\left({here}\:{with}\:{h}=\mathrm{1}\right) \\ $$$$\left({a}−{r}\right)\left({b}−{r}\right)={h}^{\mathrm{2}} \\ $$$${ab}\left(\frac{{a}+{c}}{{a}+{b}+{c}}\right)\left(\frac{{b}+{c}}{{a}+{b}+{c}}\right)={h}^{\mathrm{2}} \\ $$$$\frac{{ab}\left({a}+{c}\right)\left({b}+{c}\right)}{\left({a}+{b}+{c}\right)^{\mathrm{2}} }={h}^{\mathrm{2}} \\ $$$${ab}=\left(\mathrm{1}+\frac{{b}}{{a}+{c}}\right)\left(\mathrm{1}+\frac{{a}}{{b}+{c}}\right){h}^{\mathrm{2}} \\ $$$${A}_{\Delta{ABC}} =\left(\mathrm{1}+\frac{{b}}{{a}+{c}}\right)\left(\mathrm{1}+\frac{{a}}{{b}+{c}}\right){h}^{\mathrm{2}} \neq{unique} \\ $$
Commented by mr W last updated on 10/Jul/21
