Question

In an isosceles triangle ABC with AB = AC and BD ⊥ AC. Prove that BD² − CD² = 2CD.AD.

Answer 1

Given: ∆ABC is an isosceles triangle with AB = AC and BD ⊥ AC.

To prove: BD² − CD² = 2CD.AD

Proof: In right ∆ADB,

$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{AD}}^2+{\mathrm{BD}}^2\left(\mathrm{Pythagoras}\mathrm{Theorem}\right) \\ \Rightarrow {\mathrm{AC}}^2={\mathrm{AD}}^2+{\mathrm{BD}}^2\left(\mathrm{AB}=\mathrm{AC}\right) \\ \Rightarrow {\left(\mathrm{AD}+\mathrm{CD}\right)}^2={\mathrm{AD}}^2+{\mathrm{BD}}^2 \\ \Rightarrow {\mathrm{AD}}^2+{\mathrm{CD}}^2+2\mathrm{AD}.\mathrm{CD}={\mathrm{AD}}^2+{\mathrm{BD}}^2 \end{gathered}$$

$\Rightarrow {\mathrm{BD}}^2-{\mathrm{CD}}^2=2\mathrm{AD}.\mathrm{CD}$ (Hence Proved)