Question

In ∆ABC, seg AD ⊥ seg BC DB = 3CD. Prove that :
2AB² = 2AC² + BC²

Answer 1

It is given that,
DB = 3 CD

∴ BC = 4 CD ....(1)


According to Pythagoras theorem,
In ∆ABD

$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{AD}}^2+{\mathrm{DB}}^2 \\ \Rightarrow {\mathrm{AD}}^2={\mathrm{AB}}^2-{\mathrm{DB}}^2...\left(2\right) \end{gathered}$$

In ∆ACD

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AD}}^2+{\mathrm{CD}}^2 \\ \Rightarrow {\mathrm{AC}}^2=\left({\mathrm{AB}}^2-{\mathrm{DB}}^2\right)+{\mathrm{CD}}^2\left(\mathrm{From}\left(2\right)\right) \\ \Rightarrow {\mathrm{AC}}^2={\mathrm{AB}}^2-{\left(3\mathrm{CD}\right)}^2+{\mathrm{CD}}^2\left(\mathrm{Given}\right) \\ \Rightarrow {\mathrm{AC}}^2={\mathrm{AB}}^2-9{\mathrm{CD}}^2+{\mathrm{CD}}^2 \\ \Rightarrow {\mathrm{AC}}^2={\mathrm{AB}}^2-8{\mathrm{CD}}^2 \\ \Rightarrow {\mathrm{AB}}^2={\mathrm{AC}}^2+8{\mathrm{CD}}^2 \\ \Rightarrow {\mathrm{AB}}^2={\mathrm{AC}}^2+8{\left(\frac{\mathrm{BC}}{4}\right)}^2\left(\mathrm{From}\left(1\right)\right) \\ \Rightarrow {\mathrm{AB}}^2={\mathrm{AC}}^2+\frac{{\mathrm{BC}}^2}{2} \\ \Rightarrow 2{\mathrm{AB}}^2=2{\mathrm{AC}}^2+{\mathrm{BC}}^2 \end{gathered}$$

Hence, 2AB² = 2AC² + BC².