Question

hay friends please help me tomorrow is my maths exam

Answer 1

$\color{blue}\huge\underline\mathcal{\:HEY\:MATE}$

Here is your Theorem...

If a line is drawn parallel to one side of a triangle to intersect other two sides in distinct points, the other two sides are divided in the same ratio.
Eg in attachment.

Hope it helps you.
#Platz ✌️

Answer 2

$\bf{Answer:}$

$\bf{Theorem}$:If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other sides are divided in the same ratio.

$\Proof$:
We are given with ΔABC in which a line parallel to side BC intetsects other two sides AB and AC at D and E respectively.

we need to prove that $\frac{AD}{DB}=\frac{AE}{EC}$.

Let us join BE and CD and then draw DM perpendicular to AC and EN perpendicular to AB.

Now, ar(ADE) = $\frac{1}{2}\times{AD}\times{EN}$
Similarly ar(BDE) = $\frac{1}{2}\times{DB}\times{EN}$
ar(ADE) = $\frac{1}{1}\times{AE}\timesDM$
ar(DEC) = $\frac{1}{2}\times{EC}\timesDM$

THEREFORE, $\frac{ar(ADE)}{ar(BDE)}$

=$\frac{(1/2\times{AD}\times{EN})}{(1/2\times{DB}\timesEN)}$

=$\frac{AD}{DB}$ _______(1)

and, $\frac{ar(ADE)}{ar(DEC)}$

=$\frac{(1/2\times{AE}\times{DM})}{(1/2\times{EC}\times{DM})}$

=$\frac{AE}{EC}$ _______(2)

Note that ΔBDE and ΔDEC are on the same base DE and between the same parallels BC and DE.
So, ar(BDE)=ar(DEC) ______(3)

Therefore, from (1), (2) and (3), we have:
$\frac{AD}{DB}$=$\frac{AE}{EC}$

$\bf{Hence\:Proved}$