Question

In △ABC, AD is a median and AL ⊥ BC.
Prove that
$$\begin{gathered} \left(\mathrm{i}\right){\mathrm{AC}}^2={\mathrm{AD}}^2+\mathrm{BC}.\mathrm{DL}+{\left(\frac{\mathrm{BC}}{2}\right)}^2 \\ \left(\mathrm{ii}\right){\mathrm{AB}}^2={\mathrm{AD}}^2-\mathrm{BC}.\mathrm{DL}+{\left(\frac{\mathrm{BC}}{2}\right)}^2 \\ \left(\mathrm{iii}\right){\mathrm{AB}}^2+{\mathrm{AC}}^2=2{\mathrm{AD}}^2-\frac{1}{4}{\mathrm{BC}}^2 \end{gathered}$$

Answer 1

(a)
In right triangle ALD
Using Pythagoras theorem, we have
AL² = AD² − DL² .....(1)
Again, In right triangle ACL
Using Pythagoras theorem, we have
$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AL}}^2+{\mathrm{LC}}^2 \\ ={\mathrm{AD}}^2-{\mathrm{DL}}^2+{\left(\mathrm{DL}+\mathrm{DC}\right)}^2\left[\mathrm{Using}\left(1\right)\right] \\ ={\mathrm{AD}}^2-{\mathrm{DL}}^2+{\left(\mathrm{DL}+\frac{\mathrm{BC}}{2}\right)}^2\left[\because \mathrm{AD}\mathrm{is}\mathrm{a}\mathrm{median}\right] \\ ={\mathrm{AD}}^2-{\mathrm{DL}}^2+{\mathrm{DL}}^2+{\left(\frac{\mathrm{BC}}{2}\right)}^2+\mathrm{BC}.\mathrm{DL} \\ \therefore {\mathrm{AC}}^2={\mathrm{AD}}^2+\mathrm{BC}.\mathrm{DL}+{\left(\frac{\mathrm{BC}}{2}\right)}^2.....\left(2\right) \end{gathered}$$
(b)
In right triangle ALD
Using Pythagoras theorem, we have
AL² = AD² − DL² .....(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{AL}}^2+{\mathrm{LB}}^2 \\ ={\mathrm{AD}}^2-{\mathrm{DL}}^2+{\mathrm{LB}}^2\left[\mathrm{Using}\left(3\right)\right] \\ ={\mathrm{AD}}^2-{\mathrm{DL}}^2+{\left(\mathrm{BD}-\mathrm{DL}\right)}^2 \\ ={\mathrm{AD}}^2-{\mathrm{DL}}^2+{\left(\frac{1}{2}\mathrm{BC}-\mathrm{DL}\right)}^2 \\ ={\mathrm{AD}}^2-{\mathrm{DL}}^2+{\left(\frac{\mathrm{BC}}{2}\right)}^2-\mathrm{BC}.\mathrm{DL}+{\mathrm{DL}}^2 \\ \therefore {\mathrm{AB}}^2={\mathrm{AD}}^2-\mathrm{BC}.\mathrm{DL}+{\left(\frac{\mathrm{BC}}{2}\right)}^2.....\left(4\right) \end{gathered}$$
(c) Adding (2) and (4), we get

$$\begin{gathered} {\mathrm{AC}}^2+{\mathrm{AB}}^2={\mathrm{AD}}^2+\mathrm{BC}.\mathrm{DL}+{\left(\frac{\mathrm{BC}}{2}\right)}^2+{\mathrm{AD}}^2-\mathrm{BC}.\mathrm{DL}+{\left(\frac{\mathrm{BC}}{2}\right)}^2 \\ =2{\mathrm{AD}}^2+\frac{{\mathrm{BC}}^2}{4}+\frac{{\mathrm{BC}}^2}{4} \\ =2{\mathrm{AD}}^2+\frac{1}{2}{\mathrm{BC}}^2 \end{gathered}$$