Question

what is thales theorem and prove that​

Answer 1

Q. What is thales theorem and prove it?

Ans :-

Thales theorem :- In a right angled triangle the square of the side opposite to the right angle is equal to the sum of square of other two sides.

If ABC is a right angle $\triangle$, such that $\angle{B}$= 90° then,

$AC^2 = AB^2 + BC^2$

Now,

Given:-

ABC is a right angle $\triangle$,$\angle{B}$ = 90°

Construction :-

Draw$BD \bot AC$

Proof:-

In $\triangle{ADB} and \triangle{ABC}$

$\angle{A} = \angle {A}$ (common)

$\angle{ADB} = \angle {ABC} = 90°$

By AA similarity criteria $\triangle {ADB } \sim \triangle {ABC}$

$\therefore$ $\dfrac{AD}{AB} = \dfrac{AB}{AC}$(CPST)

$AB^2 = AD. AC$ ----eq. 1

Now in $\triangle {CDB}and \triangle {CBA}$

$\angle{c} = \angle {c}$ (common)

$\angle {CDB} = \angle{CBA} = 90°$

By AA similarity criteria $\triangle{CDB} \sim \triangle {CBA}$

$\therefore$$\dfrac{CD}{CB}= \dfrac{CB}{CA}$(CPST)

$CB^2 = CD. CA$ ------ eq. 2

Adding eq. 1 and eq. 2 we get,

$AB^2 + BC^2 = AD. AC + DC. AC$

$AB^2 + BC^2 = AC ( AD + DC)$

(AD + DC = AC)

$AB^2 + BC^2 = AC . AC$

$AB^2 + BC^2 = AC^2$ proved.

Answer 2

Answer:

Proof on Thales theorem :

If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio. Given : In ∆ABC , DE. || BC and intersects AB in D and AC in E. Provethat : AD / DB = AE / EC.