Question

state and prove Thales theorem​

Answer 1

In geometry, Thales' theorem states that if A, B, and C are distinct points on a circle where the line AC is a diameter, then the angle ∠ABC is right angle


Since OA = OB = OC, ∆OBA and ∆OBC are isosceles triangles, and by the equality of the base angles of an isosceles triangle, ∠OBC = ∠OCB and ∠BAO = ∠ABO.

Let α = ∠BAO and β = ∠OBC. The three internal angles of the ∆ABC triangle are α, (α + β), and β. Since the sum of the angles of a triangle is equal to 180°, we have

a+(a+b)=b=180°
2a+2b=180°
2(a+b)=180°
a+b= 90°

Answer 2

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

ADBD=AECE

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 PQ in points D and E, such that we get triangles A-D-E and A-E-C.

To Prove: ADBD=AECE

Construction: Join segments DC and BE

Proof:

In ΔADE and ΔBDE,

A(ΔADE)A(ΔBDE)=ADBD                 (triangles with equal heights)

In ΔADE and ΔCDE,

A(ΔADE)A(ΔCDE)=AECE                  (triangles with equal heights)

Since ΔBDE and ΔCDE have a common base DE and have the same height we can say that,

A(ΔBDE)=A(ΔCDE)

Therefore,

A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)

Therefore,

ADBD=AECE

Hence Proved.