Question

The figure shows a triangle ABC in which AD is a median and AE $\perp$ BC.
Prove that : $2A{B}^2+2A{B}^2=4A{D}^2+B{C}^2+4BC\cdot DE$

Answer 1

In ∆ ABC, AD is the median

Construction: Draw AE ⊥BC

Now since AD is the median

$\therefore BD=CD=\frac{1}{2}BC$ ....... (1)

In ∆ AED

$A{D}^2=A{E}^2+D{E}^2$ (Pythagoras theorem)

$A{E}^2=A{D}^2–D{E}^2$ ......... (2)

In ∆ AEB

$A{B}^2=A{E}^2+B{E}^2$

$=A{D}^2–D{E}^2+B{E}^2$ (from (2))

$=(BD+DE{)}^2+A{D}^2–D{E}^2$ .... $(\because BE=BD+DE)$

$=B{D}^2+D{E}^2+2BD\cdot DE+A{D}^2–D{E}^2$

$=B{D}^2+A{D}^2+2\cdot BD\cdot DE$

$={\left(\frac{1}{2}BC\right)}^2+A{D}^2+\frac{1}{2}\times 2\times BC\times DE$

$=\frac{1}{4}B{C}^2+A{D}^2+BC\times DE$ .......... (3)

$\Rightarrow A{B}^2+A{B}^2=\frac{1}{4}B{C}^2+A{D}^2+BC\cdot DE+\frac{1}{4}B{C}^2+A{D}^2+BC\cdot DE$

$\Rightarrow 2(A{B}^2+A{B}^2)=B{C}^2+4A{D}^2+4BC\cdot DE$