Question

In figure, D is the mid-point of side BC and AE$\perp$BC. If BC$=$a, AC$=$b, AB$=$C, ED$=$x, AD$=$p and AE$=$h, prove that $b^2+c^2=2p^2+\frac{a^2}{2}$.

Answer 1

Given:

$AE\perp BC$

$BD=CD$

$BC=a,AB=c,AD=p,AC=b,ED=x,AE=h$

Proof:-

In $△ABC$,

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

$\Rightarrow A{B}^2=A{E}^2+(BC-ED-CD{)}^2$

$\Rightarrow A{B}^2=A{E}^2+(BC-ED-\frac{BC}{2}{)}^2$

$\Rightarrow c^2=h^2+(\frac{a}{2}-x{)}^2$

$\Rightarrow c^2=h^2+\frac{a^2}{4}+x^2-ax$-------------(1)

In $△AEC$,

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

$\Rightarrow b^2=h^2+(\frac{BC}{2}+ED{)}^2$

$\Rightarrow b^2=h^2+(\frac{a}{2}+x{)}^2$

$\Rightarrow b^2=h^2+\frac{a^2}{4}+x^2+ax$---------------(2)

In $△AED$

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

$\Rightarrow h^2+x^2=p^2$-------------------(3)

Using Equation (3) in (1), we get

$c^2=h^2+x^2+\frac{a^2}{4}-ax$

$c^2=p^2-ax+\frac{a^2}{4}$-----------------------(4)

Using Equation(3) in (2),

$b^2=h^2+x^2+\frac{a^2}{4+ax}$

$b^2=p^2+ax+\frac{a^2}{4}$-----------------------(5)

Adding Equation (4) & (5), we get

$c^2+b^2=2p^2+\frac{a^2}{2}$