Question

proof of pythagoras theorem..... plz explain it easily ​

Answer 1

Answer:

Step-by-step explanation:triangle ABC is a right angled triangle

CA is the hypotenuse

$CA^{2}$= $AB ^{2}$+$BC^{2}$

=$\sqrt{CA}$$\sqrt{CA}$

=CA

hope you would understand

Answer 2

$\huge\sf\gray{To\;Prove}$

✭ The Pythagoras Theorem(AC² = AB² + BC²)

$\rule{110}1$

$\huge\sf\purple{Steps}$

The Pythagoras Theorem

≫ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

$\underline{\underline{\red{\sf Construction}}}$

◈ $\sf Draw \ BD \ \perp \ AC$

Proving

$\sf In \ \triangle ADB \ and \ \triangle ABC$

➝ $\sf\angle A = \angle A \: \bigg\lgroup Common\bigg\rgroup$

➝ $\sf\angle ADB = \angle ABC \bigg\lgroup \small Each \ equal \ to \ 90^{\circ}\bigg\rgroup$

$\sf\therefore \triangle ADB \sim \triangle ABC \bigg\lgroup AA \bigg\rgroup$

➢ $\sf\dfrac{AD}{AB}=\dfrac{AB}{AC}$

➢ $\sf AB^2 = AD \times AC\qquad-eq(1)$

$\sf In \ \triangle BDC \ and \ \triangle ABC$

➠ $\sf\angle C = \angle C \bigg\lgroup Common \bigg\rgroup$

➠ $\sf\angle BDC = \angle ABC \bigg\lgroup \small Each \ equal \ to \ 90^{\circ}\bigg\rgroup$

$\sf\therefore \triangle BDC \sim \triangle ABC \bigg\lgroup AA \bigg\rgroup$

»» $\sf\dfrac{DC}{BC} = \dfrac{BC}{AC}$

»» $\sf BC^2 = DC × AC\qquad -eq(2)$

Adding in eq(1) and eq(2)

➳ $\sf AB^2 + BC^2 = AD × AC + DC × AC$

➳ $\sf AB^2 + BC^2 = AC( AD + DC )$

➳ $\sf AB^2 + BC^2 = AC × AC$

➳ $\sf\orange{AC^2 = AB^2 + BC^2}$

$\sf Hence \ Prove !!$

$\rule{170}3$