Question

In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i) ${\displaystyle {\text{AC}}^2={\text{AD}}^2+\text{BC}.\text{DM}+{\left(\frac{\text{BC}}{2}\right)}^2}$

(ii) ${\displaystyle {\text{AB}}^2={\text{AD}}^2-\text{BC}.\text{DM}+{\left(\frac{\text{BC}}{2}\right)}^2}$

(iii) ${\displaystyle {\text{AC}}^2+{\text{AB}}^2=2{\text{AD}}^2+\frac{1}{2}{\text{BC}}^2}$

Answer 1

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, DC = ${\displaystyle \frac{\text{BC}}{2}}$, we obtain

${\displaystyle {\text{AD}}^2+{\left(\frac{\text{BC}}{2}\right)}^2+2\text{MD}.\left(\frac{\text{BC}}{2}\right)={\text{AC}}^2}$

${\displaystyle {\text{AD}}^2+{\left(\frac{\text{BC}}{2}\right)}^2+\text{MC}\times \text{BC}={\text{AC}}^2}$

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

= (AD² − DM²) + MB²

= (AD² − DM²) + (BD − MD)²

= AD² − DM² + BD² + MD² − 2BD × MD

= AD² + BD² − 2BD × MD

= ${\displaystyle {\text{AD}}^2+{\left(\frac{\text{BC}}{2}\right)}^2-2\left(\frac{\text{BC}}{2}\right)\times \text{MD}}$

= ${\displaystyle {\text{AD}}^2+{\left(\frac{\text{BC}}{2}\right)}^2-\text{BC}\times \text{MD}}$

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC² … (2)

Adding equations (1) and (2), we obtain

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD − DM)² + (MD + DC)² = AB² + AC²

2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²

2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC)

= AB² + AC²

= ${\displaystyle 2({\text{AM}}^2+{\text{MD}}^2)+{\left(\frac{\text{BC}}{2}\right)}^2+{\left(\frac{\text{BC}}{2}\right)}^2+2\text{MD}(\frac{-\text{BC}}{2}+\frac{\text{BC}}{2})={\text{AB}}^2+{\text{AC}}^2}$

${\displaystyle {\text{2AD}}^2+\frac{{\text{BC}}^2}{2}={\text{AB}}^2+{\text{AC}}^2}$