Question

state and proof thales theorem

Answer 1

In geometry, Thales' theorem states that "whenever an angle is drawn from the diameter of the circle to the point on its circumference, then angle formed is sure to be a perfect right angle."Â

In other words, if A,B and C are the points on a circle and AC is the diameter, then ∠∠ ABC is of 90 degree i.e., a right angle.

Thales' Theorem

While keeping the statement of theorem in mind, lets prove the theorem:

Suppose that ∠∠A = αα and ∠∠C = ββ. Then ∠∠ B = αα + ββ

Thales' Theorem Proof

Thales was at point of his understanding in geometry to believe two things:

1) The three interior angles of any triangle sum to 180 degree.

2) The two base angles of an isosceles triangle are congruent.

As we can see in above figure:

∠∠ A + ∠∠ B + ∠∠ C = 18000

αα + (αα + ββ) + ββ = 18000

Thus,

2αα + 2ββ = 18000

Giving,

αα + ββ = 9000

∠∠B = 9000

we have a right angle.

Thales Theorem Triangle

If we talk about Thales theorem triangle, we will familiar with an important truth about equiangular triangle. Thales stated this theorem as ratio of any two corresponding sides of an equiangular triangle is always same irrespective of their sizes. And this is known as BPT (Basic Proportionality Theorem).

If we draw a line parallel to any one arm of a triangle which passes through the other two arms, then other arms are divided in the same ratio.

Thales' Theorem Triangle

If DE || BC, then

AE/EC=AD/DB

Answer 2

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

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: 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.