Question Number 214859 by MATHEMATICSAM last updated on 21/Dec/24
$$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{such}\:\mathrm{that}\:\mathrm{AB}\:=\:\mathrm{AC}.\: \\ $$$$\mathrm{D}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{side}\:\mathrm{AC}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{BC}^{\mathrm{2}} \:=\:\mathrm{AC}.\mathrm{CD}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{BD}\:=\:\mathrm{BC}. \\ $$
Answered by A5T last updated on 21/Dec/24
Commented by A5T last updated on 21/Dec/24
$${BC}^{\mathrm{2}} ={CA}×{CD}\Rightarrow{CD}=\frac{{BC}^{\mathrm{2}} }{{CA}} \\ $$$$\angle{BCD}=\theta\Rightarrow{AB}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{AC}^{\mathrm{2}} −\mathrm{2}{BC}×{ACcos}\theta \\ $$$$\Rightarrow{cos}\theta=\frac{{BC}}{\mathrm{2}{AC}} \\ $$$${BD}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{CD}^{\mathrm{2}} −\mathrm{2}{BC}×{CDcos}\theta \\ $$$$\Rightarrow{BD}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{CD}^{\mathrm{2}} −{CD}×\frac{{BC}^{\mathrm{2}} }{{CA}} \\ $$$$\Rightarrow{BD}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{CD}^{\mathrm{2}} −{CD}^{\mathrm{2}} ={BC}^{\mathrm{2}} \\ $$$$\Rightarrow{BD}={BC} \\ $$
Commented by MATHEMATICSAM last updated on 22/Dec/24
$$\mathrm{Can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{proved}\:\mathrm{by}\:\mathrm{pythagoras} \\ $$$$\mathrm{identity}\:\mathrm{without}\:\mathrm{cosine}\:\mathrm{law}\:\mathrm{or}\:\mathrm{something}? \\ $$
Commented by A5T last updated on 22/Dec/24
$${I}\:{don}'{t}\:{really}\:{know}\:{about}\:{that},\:{but}\:{cosine}\:{rule} \\ $$$${could}\:{be}\:{seen}\:{as}\:{a}\:{generalized}\:{form}\:{of}\: \\ $$$${Pythagoras}'\:{theorem}\left({a}\:{special}\:{case}\:{when}\:\theta=\mathrm{90}°\right) \\ $$
Answered by mr W last updated on 22/Dec/24
Commented by mr W last updated on 22/Dec/24
$${BC}^{\mathrm{2}} ={AC}×{CD} \\ $$$$\Rightarrow\frac{{BC}}{{AC}}=\frac{{CD}}{{BC}}\: \\ $$$$\Rightarrow\Delta{ABC}\sim\Delta{BDC} \\ $$$$\Rightarrow\angle{BDC}=\angle{ABC}=\angle{BCD} \\ $$$$\Rightarrow{BD}={BC} \\ $$