Question

In an equilateral $\Delta ABC,$ D is a point on side BC such that BD $=\frac{1}{3}$BC. Prove that $9(AD{)}^2=7(AB{)}^2$.
OR
Prove that, in a right triangle, the square on the hyporenuse is equal to the sum of the squares on the other two sides.

Answer 1

Given, ABC is an equilateral triangle and D is a point on BC such that BD $=\frac{1}{3}$ BC.
To prove :
$9A{D}^2=7A{B}^2$
Construction : Draw AE $\perp$ BC
Proof : BD = $\frac{1}{3}$ BC ......(i)(Given)
AE $\perp$ BC
We know that perpendicular from a vertex of equilateral triangle to the base divides base in two equal parts.
$\therefore BE=EC=\frac{1}{2}BC$. .........(ii)
In $\Delta AEB$,
$A{D}^2=A{E}^2+D{E}^2$ (Pythagoras theorem)
or
$A{E}^2=A{D}^2-D{E}^2$ ......(iii)
= $A{D}^2-D{E}^2+\left(\frac{1}{2}B{C}^2\right)$ [From (ii) and (iii)]
$=A{D}^2-(BE-BD{)}^2+\frac{1}{4}B{C}^2$
$=A{D}^2-B{E}^2-B{D}^2+2.BE.BD+\frac{1}{4}B{C}^2$
$=A{D}^2-{\left(\frac{1}{2}BC\right)}^2-{\left(\frac{1}{3}BC\right)}^2+2.\frac{1}{2}BC.\frac{1}{3}BC+\frac{1}{4}B{C}^2$
$A{B}^2=A{D}^2-\frac{1}{9}B{C}^2+\frac{1}{3}B{C}^2$
$\Rightarrow A{B}^2=A{D}^2+\frac{2}{9}B{C}^2$
$\Rightarrow A{B}^2=A{D}^2+\frac{2}{9}A{B}^2$ $(\because BC=AB)$
$\Rightarrow A{B}^2-\frac{2}{9}A{B}^2=A{D}^2$
$\Rightarrow \frac{7}{9}A{B}^2=A{D}^2$
$\Rightarrow 7A{B}^2=9A{D}^2$
Or $9(AD{)}^2=7(AB{)}^2$ Hence Proved.
OR


Given : $\Delta ABC$ is a right angle triangle, right angled at A.
To prove : $B{C}^2=A{B}^2+A{C}^2$
Construction : Draw AD $\perp$ BC.
Proof : In $\Delta ADB$ and
$\Delta$ BAC,
$\angle B=\angle B$ (Common)
$\angle ADB=\angle BAC$ (Each ${90}^o$)
$\therefore \Delta ADB\sim \Delta BAC$ (By AA similarity axiom)
$\therefore \frac{AB}{BC}=\frac{BD}{AB}$
$A{B}^2=BC\times BD$
Similarly,
$\Delta ADC-\Delta CAB$
$\frac{AC}{BC}=\frac{DC}{AC}$
$A{C}^2=BC\times DC$ ......(ii)
On adding eq. (i) and (ii)
$A{B}^2+A{C}^2=BC\times BD+BC\times CD$
$=BC(BD+CD)$
$=BC\times BC$
$A{B}^2+A{C}^2=B{C}^2$
$\Rightarrow B{C}^2=A{B}^2+A{C}^2$
Hence Proved.