THALES THEOREM

Post Summarization:

● Thales Theorem and its proof.

What is Thales's Theorem?

If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally.

Given data,

In ∆ABC, DE || BC

Aim to prove: AD/DB = AE/EC

First of all, join 'D' with 'C' and 'E' with 'B' with straight lines.

Next, draw 'EL' perpendicular (90°) to 'AB' and 'DN' perpendicular (90°) to 'AC'.

Now, as 'EL' is perpendicular (90°) to 'AB', therefore, 'EL' is the height of ∆AEL and ∆DEL.

Again, ∆DEL and ∆BDE have a common apex point 'E', and the base of both triangles are on the same plane,

Therefore, 'EL' is the height of ∆BDE.

Similarly, 'DN' is the height of ∆CDE.

As we know, the area of a triangle = 1/2×b×h

Where, 

b = Base of the triangle, and 

h = Height of triangle.

Therefore, 

AREA of ∆ADE / AREA of ∆BDE

= (1/2×AD×EL) / (1/2×DB×EL)

= AD / DB    ....................(i)

Similarly,

AREA of ∆ADE / AREA of ∆CDE

= (1/2×AE×DN) / (1/2×EC×DN)

= AE / EC    ....................(ii)

[ As, ∆BDE and ∆CDE are congruent triangles.

Therefore, AREA ∆BDE = AREA ∆CDE ]

Now, from equation no. (i) and (ii), we get,

AD/DB = AE/EC (Proved).