Question

In fig., $AD$ is a median of a triangle $ABC$ and $AM\perp BC$. Prove that:

(i) $A{C}^2=A{D}^2+BC.DM+{\left(\frac{BC}{2}\right)}^2$

(ii) $A{B}^2=A{D}^{2}BC.DM+{\left(\frac{BC}{2}\right)}^2$

(iii) $A{C}^2+A{B}^2=2A{D}^2+\frac{1}{2}B{C}^2$

Answer 1

It is given that

$\angle AMD={90}^0$

Referring to the figure, we can say that

$\angle ADM<{90}^0$ and $\angle ADC>{90}^0$

Now,

(i)

To prove:

$A{C}^2=A{D}^2+BC.DM+{\left(\frac{BC}{2}\right)}^2$

In $\Delta ADC,\angle ADC$ ia an obtuse angle.

$\therefore A{C}^2=A{d}^2+D{C}^2+2DC.DM$

$\Rightarrow A{C}^2=A{D}^2+{\left(\frac{BC}{2}\right)}^2+2.\frac{BC}{2}.DM$

$\Rightarrow A{C}^2=A{D}^2+BC.DM+{\left(\frac{BC}{2}\right)}^2$

(ii)

To prove:

$A{B}^2=A{D}^2-BC.DM+{\left(\frac{BC}{2}\right)}^2$

In $\Delta ABD,\angle ADM$ is an obtuse angle.

$\therefore A{B}^2=A{D}^2+B{D}^2-2BD.DM$

$\Rightarrow A{B}^2=A{D}^2+{\left(\frac{BC}{2}\right)}^2-2.\frac{BC}{2}.DM$

$\Rightarrow A{B}^2=A{D}^2-BC.DM+{\left(\frac{BC}{2}\right)}^2$

(iii)

To prove:

$A{C}^2+A{B}^2=2A{D}^2+\frac{1}{2}B{C}^2$

From the result of (i) and (ii), adding those, we get

$A{C}^2+A{B}^2=2A{D}^2+\frac{1}{2}B{C}^2$