Question

prove basic propornality theoram​

Answer 1

Basic Proportionality Theorem. (BPT)

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

Proof for BPT, or Basic Proportionality Theorem.

Given:

In ΔABC,

DE║BC

To Prove:

$\sf \dfrac{AD}{AB} = \dfrac{AE}{EC}$

Construction:

Construct a line PE ⊥ AD, and DQ ⊥ AC.

Join BE and CD.

Proof:

ar(ADE) = ¹/₂ × base × height.

ar(ADE) = ¹/₂ × AD × PE → Eq(1)

Area of ADE is also equal to,

ar(ADE) = ¹/₂ × AE × DQ → Eq(2)

In ΔBDE,

ar(BDE) =  ¹/₂ × base × height.

ar(BDE) =  ¹/₂ × BD × PE → Eq(3)

In ΔCED

ar(CED) =  ¹/₂ × CE × DQ → Eq(4)

On Dividing equations (1) and (3) we get,

$\Longrightarrow \sf \dfrac{ar(ADE)}{ar(BDE)} = \dfrac{\frac{1}{2} \times AD \times PE}{\frac{1}{2} \times BD \times PE}$

$\Longrightarrow \sf \dfrac{ar(ADE)}{ar(BDE)} = \dfrac{AD}{BD} \longrightarrow Eq(5)$

On Dividing equations (2) and (4) we get,

$\Longrightarrow \sf \dfrac{ar(ADE)}{ar(CED)} = \dfrac{\frac{1}{2} \times AE \times DQ}{\frac{1}{2} \times CE \times DQ}$

$\Longrightarrow \sf \dfrac{ar(ADE)}{ar(CED)} = \dfrac{AE}{CE} \longrightarrow Eq(6)$

We also know that Triangles that lie between the same parallels and have the same base are equal in area. Hence,

ar(BDE) = ar(CED) → Eq(7)

Substituting Equation 7 in Equation 5 we get,

$\Longrightarrow \sf \dfrac{ar(ADE)}{ar(CED)} = \dfrac{AD}{BD} \longrightarrow Eq(8)$

From Equations 6 and 8, we can say that,

$\Longrightarrow \sf \dfrac{AE}{CE} =\dfrac{AD}{BD}$

Hence Proved.

Answer 2

Answer:

By Basic proportional theoram

AD  =  AE

DB      CE

this is the method