Question

In the given figure, D is the midpoint of side BC and AE $\perp$ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that

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

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

(iii) $(b^2+c^2)=2p^2+\frac{1}{2}{a}^2$

(iv) $(b^2-c^2)=2ax$

Answer 1

Given: D is the midpoint of side BC, AE ⊥ BC,
BC = a, AC = b, AB = c, ED = x, AD = p and AE = h

In ΔAEC, $\angle AEC={90}^o$

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

⇒ $p^2=h^2+x^2$

(i) In ΔAEC, $\angle AEC={90}^o$

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

Therefore, $b^2=p^2+ax+\frac{a^2}{4}$——- (1)

(ii) In ΔABE, $\angle ABE={90}^o$

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

⇒$c^2=h^2+(\frac{a}{2}-x{)}^2$...(2)

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

= $p2-ax+\frac{a^2}{4}$

Hence, $c^2=p^2-ax+\frac{a^2}{4}$

(iii) Adding (1) and (2), we get:

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

$(b^2+c^2)=2p^2+\frac{1}{2}{a}^2$

(iv)
Subtracting (2) from (1), we get:

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

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

$(b^2-c^2)=2ax$