Question

In the given figure, ∆ABC is an obtuse triangle, obtuse-angles at B. If AD ⊥ CB, Prove that AC² = AB² + BC² + 2BC ⋅ BD.

Answer 1

Applying Pythagoras theorem in right-angled triangle ADC, we get:

$$\begin{gathered} A{C}^2=A{D}^2+D{C}^2 \\ \Rightarrow A{C}^2-D{C}^2=A{D}^2 \\ \Rightarrow A{D}^2=A{C}^2-D{C}^2...\left(1\right) \end{gathered}$$

Applying Pythagoras theorem in right-angled triangle ADB, we get:

$$\begin{gathered} A{B}^2=A{D}^2+D{B}^2 \\ \Rightarrow A{B}^2-D{B}^2=A{D}^2 \\ \Rightarrow A{D}^2=A{B}^2-D{B}^2...\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{From}\mathrm{equation}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{have}: \\ A{C}^2-D{C}^2=A{B}^2-D{B}^2 \\ \Rightarrow A{C}^2=A{B}^2+D{C}^2-D{B}^2 \\ \Rightarrow A{C}^2=A{B}^2+(DB+BC{)}^2-D{B}^2\left(\because DB+BC=DC\right) \\ \Rightarrow A{C}^2=A{B}^2+D{B}^2+B{C}^2+2DB.BC-D{B}^2 \\ \Rightarrow A{C}^2=A{B}^2+B{C}^2+2BC.BD \end{gathered}$$

This completes the proof.