Question

In the given figure, D is the midpoint of side BC and AE ⊥ 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

(i)
In right-angled triangle AEC, applying Pythagoras theorem, we have:

$$\begin{gathered} A{C}^2=A{E}^2+E{C}^2 \\ \Rightarrow b^2=h^2+{\left(x+\frac{a}{2}\right)}^2=h^2+x^2+\frac{a^2}{4}+ax...\left(i\right) \\ \mathrm{In}\mathrm{right}-\mathrm{angled}\mathrm{triangle}\mathrm{AED},\mathrm{we}\mathrm{have}: \\ A{D}^2=A{E}^2+E{D}^2 \\ \Rightarrow p^2=h^2+x^2...\left(ii\right) \\ \mathrm{Therefore}, \\ \mathrm{from}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right), \\ {b}^2=p^2+ax+\frac{a^2}{4} \end{gathered}$$

(ii)
In right-angled triangle AEB, applying Pythagoras theorem, we have:
$$\begin{gathered} A{B}^2=A{E}^2+E{B}^2 \\ \Rightarrow c^2=h^2+{\left(\frac{a}{2}-x\right)}^2(\because BD=\frac{a}{2}andBE=BD-x) \\ \Rightarrow c^2=h^2+x^2-ax+\frac{a^2}{4}(\because h^2+x^2=p^2) \\ \Rightarrow c^2=p^2-ax+\frac{a^2}{4} \end{gathered}$$

(iii)

Adding (i) and (ii), we get:
$$\begin{gathered} \Rightarrow b^2+c^2=p^2+ax+\frac{a^2}{4}+p^2-ax+\frac{a^2}{4} \\ =2p^2+ax-ax+\frac{a^2+a^2}{4} \\ =2p^2+\frac{a^2}{2} \end{gathered}$$

(iv)
Subtracting (ii) from (i), we get:
$$\begin{gathered} b^2-c^2=p^2+ax+\frac{a^2}{4}-(p^2-ax+\frac{a^2}{4}) \\ =p^2-p^2+ax+ax+\frac{a^2}{4}-\frac{a^2}{4} \\ =2ax \end{gathered}$$