Question

Seg AD is the median of $△ABC$ and AM $\perp$ BC.

Prove that:

i) $A{C}^2=A{D}^2+BC\times DM+\frac{B{C}^2}{4}$

ii) $A{B}^2=A{D}^2-BC\times DM+\frac{B{C}^2}{4}$

Answer 1

i) $In\Delta AMC,$

${AC}^2={AM}^2+{CM}^2$ [By Pythagoras Theorem]....(i)

$In\Delta AMD,$

${AD}^2={AM}^2+{DM}^2$ [By Pythagoras Theorem]....(ii)

$\therefore {AM}^2={AD}^2-{DM}^2$

$\Rightarrow {AC}^2=[{AD}^2-{DM}^2]+{CM}^2$

${AC}^2={AD}^2-{DM}^2+{(DC+DM)}^2$

$={AD}^2-{DM}^2+{DC}^2+2DC.DM+{DM}^2$

$={AD}^2+{DC}^2+2DC.DM$

$={AD}^2+{\left(\frac{BC}{2}\right)}^2+2\frac{BC}{2}DM$

${AC}^2={AD}^2+BC.DM+\frac{{BC}^2}{4}$

ii) From $△ABM$

${AB}^2={AM}^2+{BM}^2$

From $△AMD,$

${AM}^2={AD}^2-{DM}^2$

$\Rightarrow {AB}^2=[{AD}^2-{DM}^2]+{BM}^2$

$\therefore {BM}^2=(BC-DC-DM{)}^2$

${(a-b-c)}^2=a^2+b^2+c^2-2ab+2bc-2ac$

${AB}^2={AD}^2-{DM}^2+[{BC}^2+{DC}^2-2(BC\times DC)+2(DC\times DM)-2(BC\times DM\left)\right]$

Put $DC=\frac{DC}{2}$

$\therefore {AB}^2={AD}^2+{BC}^2+\frac{{BC}^2}{4}-2\frac{{BC}^2}{2}+2(\frac{BC}{2}\times DM)-2(BC\times DM)$

$\Rightarrow {AB}^2={AD}^2+\frac{{BC}^2}{4}-(BC\times DM)$