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
$${BC}^{\mathrm{2}} ={CA}×{CD}\Rightarrow{CD}=\frac{{BC}^{\mathrm{2}} }{{CA}} \\ $$$$\angle{BCD}=\theta\Rightarrow{AB}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{AC}^{\mathrm{2}} −\mathrm{2}{BC}×{ACcos}\theta \\ $$$$\Rightarrow{cos}\theta=\frac{{BC}}{\mathrm{2}{AC}} \\ $$$${BD}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{CD}^{\mathrm{2}} −\mathrm{2}{BC}×{CDcos}\theta \\ $$$$\Rightarrow{BD}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{CD}^{\mathrm{2}} −{CD}×\frac{{BC}^{\mathrm{2}} }{{CA}} \\ $$$$\Rightarrow{BD}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{CD}^{\mathrm{2}} −{CD}^{\mathrm{2}} ={BC}^{\mathrm{2}} \\ $$$$\Rightarrow{BD}={BC} \\ $$