Question

write the statement of phthagorous therom and prove it​

Answer 1

TO PROVE :-

• The Pythagoras theorem (AC² = AB² + BC²)

CONSTRUCTION :-

• Draw BD ⟂ AC.

PROOF :-

Pythagoras theorem :- the Pythagoras theorem states that in every right angled ∆ the square of hypotenuse is is equal to the sum of square of perpendicular and base.

In ∆ ABD and ∆ ABC,

$\\ : \implies \displaystyle \sf \angle A = \angle A \ \ \ \ \ \ \ \ \ \ \bigg \lgroup \displaystyle \sf Common \bigg \rgroup$

$\\ : \implies \displaystyle \sf \angle D = \angle B \ \ \ \ \ \ \ \ \ \ \bigg \lgroup \displaystyle \sf Perpendiculars \bigg \rgroup$

$\therefore \underline {\displaystyle \sf By \ AA \: Similarity \: \triangle ABD \: \sim\triangle ABC.}$

As we know the theorem :- sides of similar triangle are in the same ratio.

$\\ : \implies \displaystyle \sf AD\ratio \: AB= AB \ratio BC$

$\\ : \implies \displaystyle \sf \dfrac{AD}{AB} = \dfrac{AB}{AC}$

$: \implies \displaystyle \sf AD \times AC = AB \times AB$

$: \implies \displaystyle \sf AD \times AC = AB ^{2} \: \: \: \: \: \: \: \: \: \bigg \lgroup \displaystyle \sf Equation \ 1 \bigg \rgroup$

Now in ∆ BDC and ∆ABC,

$\\ : \implies \displaystyle \sf \angle C = \angle C \ \ \ \ \ \ \ \ \ \ \bigg \lgroup \displaystyle \sf Common \bigg \rgroup$

$\\ : \implies \displaystyle \sf \angle D = \angle B \ \ \ \ \ \ \ \ \ \ \bigg \lgroup \displaystyle \sf Perpendiculars \bigg \rgroup$

$\therefore \underline {\displaystyle \sf By \ AA \ Similarity \ \triangle BDC \ \sim \triangle ABC.}$

As we know the theorem :- sides of similar triangle are in the same ratio.

$\\ : \implies \displaystyle \sf CD\ratio \ BC = BC \ratio AC$

$: \implies \displaystyle \sf \dfrac{ CD} {BC} = \dfrac{BC}{ AC}$

$: \implies \displaystyle \sf AC \times CD = BC \times BC$

$: \implies \displaystyle \sf AC \times CD = BC^{2} \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \displaystyle \sf Equation \ 1 \bigg \rgroup$

_____________________

$\\ \dashrightarrow \bigg \lgroup \displaystyle \sf Equation \ 1 \bigg \rgroup + \bigg \lgroup \displaystyle \sf Equation \ 2 \bigg \rgroup$

$\dashrightarrow \displaystyle \sf AD \times AC + AC \times CD = AB^{2} + BC^{2}$

$\dashrightarrow \displaystyle \sf \: AC( CD + AD) = AB^{2} + BC^{2} \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \displaystyle \sf Common \ AC \bigg \rgroup$

$\dashrightarrow \displaystyle \sf \: AC ^{2} = AB^{2} + BC^{2} \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \displaystyle \sf AC = AD + CD\bigg \rgroup$

HENCE PROVED.

Answer 2

Answer:

Step-by-step explanation:

Pythagoras theorem states that “ In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

The sides of the right-angled triangle are called base, perpendicular and hypotenuse . Proof: ... ∠BAD = ∠BAC i.e. ∠A is common in both triangles.