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Question Number 214859 by MATHEMATICSAM last updated on 21/Dec/24

$$\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{such}\mathrm{that}\mathrm{AB}=\mathrm{AC}.$$$$\mathrm{D}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{on}\mathrm{side}\mathrm{AC}\mathrm{such}$$$$\mathrm{that}{\mathrm{BC}}^2=\mathrm{AC}.\mathrm{CD}.\mathrm{Prove}\mathrm{that}$$$$\mathrm{BD}=\mathrm{BC}.$$

Answered by A5T last updated on 21/Dec/24

Commented by A5T last updated on 21/Dec/24

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

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

$$I{don}^{\prime }treallyknowaboutthat,butcosinerule$$$$couldbeseenasageneralizedformof$$$${Pythagoras}^{\prime }theorem(aspecialcasewhen\theta =90°)$$

Answered by mr W last updated on 22/Dec/24

Commented by mr W last updated on 22/Dec/24

$${BC}^2=AC\times CD$$$$\Rightarrow \frac{BC}{AC}=\frac{CD}{BC}$$$$\Rightarrow \mathrm{\Delta }ABC\sim \mathrm{\Delta }BDC$$$$\Rightarrow \mathrm{\angle }BDC=\mathrm{\angle }ABC=\mathrm{\angle }BCD$$$$\Rightarrow BD=BC$$

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