Question

Explain basic proportional theorem

Answer 1

Answer:

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion

Step-by-step explanation:

Line DE intersects sides AB and AC in points D and E respectively.

To Prove: ADBD=AECE

Construction: Draw EF ⟂ AD and DG⟂ AE and join the segments BE and CD.

Proof: 

Area of Triangle= ½ × base × height

In ΔADE and ΔBDE,

Ar(ADE)Ar(DBE)=12×AD×EF12×DB×EF=ADDB(1)

In ΔADE and ΔCDE,

Ar(ADE)Ar(ECD)=12×AE×DG12×EC×DG=AEEC(2)

Note that ΔDBE and ΔECD have a common base DE and lie between the same parallels DE and BC. Also, we know that triangles having the same base and lying between the same parallels are equal in area.

So, we can say that

Ar(ΔDBE)=Ar(ΔECD)

Therefore,

A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)

Therefore,

ADBD=AECE

Hence Proved.

Answer 2

Step-by-step explanation:

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion. Let’s not stop at the statement, we need to find a proof that its true. So shall we begin?

Introduction

Basic Proportionality Theorem was first stated by Thales, a Greek mathematician. Hence it is also known as Thales Theorem. Thales first initiated and formulated the Theoretical Study of Geometry to make astronomy a more exact science. What is this theorem that Thales found important for his study of astronomy? Let us find it out.

Basic Proportionality Theorem (can be abbreviated as BPT) states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Basic Proportionality Theorem

In the figure alongside, if we consider DE is parallel to BC, then according to the theorem,

\(\frac{AD}{BD} = \frac{AE}{CE} \)

In the figure alongside, if we consider DE is parallel to BC, then according to the theorem,

\(\frac{AD}{BD} = \frac{AE}{CE} \)

Let’s not stop at the statement, we need to find a proof that its true. So shall we begin?

PROOF OF BPT

Given: In ΔABC, DE is parallel to BC

Line DE intersects sides AB and AC in points D and E respectively.

To Prove: \( \frac{AD}{BD} = \frac{AE}{CE} \)

Construction: Draw EF ⟂ AD and DG⟂ AE and join the segments BE and CD.

Proof:

Area of Triangle= ½ × base × height

In ΔADE and ΔBDE,

$$\frac{Ar(ADE)}{Ar(DBE)} =\frac{\frac{1}{2}\times AD \times EF}{\frac{1}{2}\times DB \times EF} = \frac{AD}{DB} \hspace{1.4cm} (1)$$

In ΔADE and ΔCDE,

$$\frac{Ar(ADE)}{Ar(ECD)} =\frac{\frac{1}{2}\times AE \times DG}{\frac{1}{2}\times EC \times DG} = \frac{AE}{EC} \hspace{1.4cm} (2)$$

Note that ΔDBE and ΔECD have a common base DE and lie between the same parallels DE and BC. Also, we know that triangles having the same base and lying between the same parallels are equal in area.

So, we can say that

$$Ar(\Delta DBE) = Ar(\Delta ECD)$$

Therefore,

\( \frac{A(ΔADE)}{A(ΔBDE)} = \frac{A(ΔADE)}{A(ΔCDE)} \)

Therefore,

\(\frac{AD}{BD} = \frac{AE}{CE} \)

Hence Proved.

The BPT also has a converse which states, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

(Note: A converse of any theorem is just a reverse of the original theorem, just like we have active and passive voices in English.

hope it's helpful for you ☺️