Question

State and prove BPT.

Answer 1

Statement : If a line passing through two sides of triangle then it os parallel to third side then divides other two sides in same ratio.

In $\Delta ADE$ (Baes AD)

Area of triangle = $\frac{1}{2}.AD.ME$ _______ (1)

Again in $\Delta ADE$ (Base AE)

$ar.\left(\Delta ADE\right)=\frac{1}{2}.AE.ND$ ________ (2)

In $\Delta BDE$ (Base BD)

$ar.\left(\Delta BDE\right)=\frac{1}{2}.DB.ME$ ________ (3)

In $\Delta DEC$ (Base EC)

$ar.\left(\Delta DEC\right)=\frac{1}{2}.EC.ND$ ________ (4)

Equation (1) & (3)

$\frac{ar.\Delta ADE}{ar.\Delta BDE}=\frac{\frac{1}{2}.AD.ME}{\frac{1}{2}.DB.ME}$

$\frac{ar\Delta ADE}{ar.\Delta BDE}=\frac{AD}{DB}$ _______ (5)

Now equation (2) & (4)

$\frac{ar.\left(\Delta ADE\right)}{ar.\left(\Delta DEC\right)}=\frac{\frac{1}{2}.AE.ND}{\frac{1}{2}.EC.ND}$

$\frac{ar.\left(\Delta ADE\right)}{\Delta DEC}=\frac{AE}{EC}$ _______ (6)

By theorem,

$ar.\left(\Delta BDE\right)=ar.\left(\Delta DEC\right)$

$\frac{ar.\left(\Delta ADE\right)}{ar.\left(\Delta BDC\right)}=\frac{ar.\left(\Delta ADE\right)}{ar.\left(\Delta DEC\right)}=\frac{AD}{BD}$

$\frac{ar.\left(\Delta ADE\right)}{ar.\left(\Delta DEC\right)}=\frac{AD}{BD}$ ________ (7)

By equation (6) & (7)

There L.H.S is same and R.H.S is same.

$\frac{AE}{EC}=\frac{AD}{BD}$

$\therefore \frac{AD}{DB}=\frac{AE}{EC}$