Question

state Thales theorem​

Answer 1

answer= in geometry Thales theorem state that if B and C are distinct points on a circle where the line AC is a diameter then the angle ABC is a right angle. Thales theorem is a special case of the inscribed angle theorem and is mentioned and proved as a part of the 31st proposition,in the third book of Euclid's elements .it is generally attributed to Thales of milatus who is said to have offered an ox (probably to the god Apollo) as the sacrifice of the thanksgiving for the discovery but some time it is attributed to Pythagoras.......

hope it will help you please mark the answer as brainlist and don't forget to follow me OK thanks

Answer 2

Hi Mate,

The Thales Theorem (or Basic Proportionality Theorem) is as follows:

"Is a line is drawn parallel to one side of a triangle to intersect other two sides in distinct points, the other two sides are divided in the same ratio"

Proof: we are we are given a triangle ABC in which a line parallel to dude BC intersects other two sides AB and AC at D and E respectively.

We need to prove that

$\frac{ad}{db} = \frac{ae}{ec}$

Let us join BE and CD and then draw DM perpendicular to AC and EN perpendicular to AB.

$now \: area \: of \: triange \: ADE \: ( = \frac{1}{2} base \: \times height)$

So, are(ADE) = ½ AD × EN

Similarly, ar(BDE) = ½ DB × EN

ar(ADE) = ½ AE × DM and ar(DEC) = ½ EC × DM

Therefore,

$ar(ade) \: = \frac{\frac{1}{2} \: ad \: \times en}{\frac{1}{2} \: db \: \times en} \: \: = \frac{ad}{db} \\ \\ and \: \: \: \: \: \: \: \frac{ar(ade)}{ar(dec)} = \frac{\frac{1}{2} \: ae \: \times dm}{\frac{1}{2} \: ec \: \times dm}$

So, ar(BDE) = ar(DEC)

Therefore,

$\frac{ad}{db} = \frac{ae}{ec}$

HENCE, PROVED

Hope this answer will help you...(◔‿◔)