Basic Proportionality Theorem & It's Proof - Goyanka Maths Study - GMS - Learning Simply
Students' favourite free learning app with LIVE online classes, instant doubt resolution, unlimited practice for classes 6-12, personalized study app for Maths, Science, Social Studies, video e-learning, online tutorial, and more. Join Telegram

# Basic Proportionality Theorem & It's Proof - Goyanka Maths Study

Scroll Down and click on Go to Link for destination

# Hint: To prove this theorem first we will join BE and CD. Then draw a line EL perpendicular to AB and line DM perpendicular to AC. Now we will find the ratio of area of Δ$\mathrm{\Delta }$ADE to Δ$\mathrm{\Delta }$DBE and ratio of area of Δ$\mathrm{\Delta }$ADE to Δ$\mathrm{\Delta }$ECD. Comparing the ratios we will get the final answer.

Now, $\mathrm{\Delta }DBE$ and $\mathrm{\Delta }ECD$ being on the same base DE and between the same parallels DE and BC, we have,$ar\left(\mathrm{\Delta }DBE\right)=ar\left(\mathrm{\Delta }ECD\right)$ then we say that the basic proportionality theorem is proved.

Basic proportionality theorem:

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

Given:
$\mathrm{\Delta }ABC$in which $DE\parallel BC$and DE intersects AB and AC at D and E respectively.

To prove that:

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

Construction:Join BE and CD.
Draw $EL\mathrm{\perp }AB$and $DM\mathrm{\perp }AC$

Proof:
We have the
$ar\left(\mathrm{\Delta }ADE\right)=\frac{1}{2}×AD×EL$
$ar\left(\mathrm{\Delta }DBE\right)=\frac{1}{2}×DB×EL$
Therefore the ratio of these two is $\frac{ar\left(\mathrm{\Delta }ADE\right)}{ar\left(\mathrm{\Delta }DBE\right)}=\frac{AD}{DB}$. . . . . . . . . . . . . . (1)
Similarly,
$ar\left(\mathrm{\Delta }ADE\right)=ar\left(\mathrm{\Delta }ADE\right)=\frac{1}{2}×AE×DM$
$ar\left(\mathrm{\Delta }ECD\right)=\frac{1}{2}×EC×DM$
Therefore the ratio of these two is $\frac{ar\left(\mathrm{\Delta }ADE\right)}{ar\left(\mathrm{\Delta }ECD\right)}=\frac{AE}{EC}$. . . . . . . . . . . .. . . (2)

Now, $\mathrm{\Delta }DBE$ and $\mathrm{\Delta }ECD$ being on the same base DE and between the same parallels DE and BC, we have,
$ar\left(\mathrm{\Delta }DBE\right)=ar\left(\mathrm{\Delta }ECD\right)$. . . . . . . . . . . (3)
From equations 1, 2, 3 we can conclude that
$\frac{AD}{DB}=\frac{AE}{EC}$
Hence we can say that the basic proportionality theorem is proved.

Note: The formula for area of the triangle is given by $\frac{1}{2}×b×h$where b, h are base and height respectively. If two triangles are on the same base and between the same parallels then the area of those two triangles are equal.

At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio…