Solve this

Solution:

(a)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2                 .....(1)
Again, In right triangle ACL
Using Pythagoras theorem, we have

$\mathrm{AC}^{2}=\mathrm{AL}^{2}+\mathrm{LC}^{2}$

$=\mathrm{AD}^{2}-\mathrm{DL}^{2}+(\mathrm{DL}+\mathrm{DC})^{2} \quad[\mathrm{Using}(1)]$

$=\mathrm{AD}^{2}-\mathrm{DL}^{2}+\left(\mathrm{DL}+\frac{\mathrm{BC}}{2}\right)^{2} \quad[\because \mathrm{AD}$ is a median $]$

$=\mathrm{AD}^{2}-\mathrm{DL}^{2}+\mathrm{DL}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}+\mathrm{BC} \cdot \mathrm{DL}$

$\therefore \mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BC} \cdot \mathrm{DL}+\left(\frac{\mathrm{BC}}{2}\right)^{2} \quad \ldots(2)$

(b)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2                 .....(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have

$\mathrm{AB}^{2}=\mathrm{AL}^{2}+\mathrm{LB}^{2}$

$=\mathrm{AD}^{2}-\mathrm{DL}^{2}+\mathrm{LB}^{2} \quad[\mathrm{Using}(3)]$

$=\mathrm{AD}^{2}-\mathrm{DL}^{2}+(\mathrm{BD}-\mathrm{DL})^{2}$

$=\mathrm{AD}^{2}-\mathrm{DL}^{2}+\left(\frac{1}{2} \mathrm{BC}-\mathrm{DL}\right)^{2}$

$=\mathrm{AD}^{2}-\mathrm{DL}^{2}+\left(\frac{1}{2} \mathrm{BC}-\mathrm{DL}\right)^{2}$

$=\mathrm{AD}^{2}-\mathrm{DL}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}-\mathrm{BC} \cdot \mathrm{DL}+\mathrm{DL}^{2}$

$\therefore \mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \cdot \mathrm{DL}+\left(\frac{\mathrm{BC}}{2}\right)^{2} \quad \ldots(4)$

(c) Adding (2) and (4), we get

$\mathrm{AC}^{2}+\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BC} \cdot \mathrm{DL}+\left(\frac{\mathrm{BC}}{2}\right)^{2}+\mathrm{AD}^{2}-\mathrm{BC} \cdot \mathrm{DL}+\left(\frac{\mathrm{BC}}{2}\right)^{2}$

$=2 \mathrm{AD}^{2}+\frac{\mathrm{BC}^{2}}{4}+\frac{\mathrm{BC}^{2}}{4}$

$=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$