Question

prove that in a right angle triangle , the square of the hypotenuse is equal to the sum of the square of the other two sides.

The perpendicular from vertex A on the side BC of a ∆ABC interests BC at point D such that DB = 3CD .prove that 2AB² = 2AC² + BC²

solve plzz......

Answer 1

here is ur answer for first question.

2nd attachment is ur 2nd answer

Answer 2

Hello , dear friend..
here is Ur answer ............

Solution.

Given : A right triangle ABC , right angled at B.

To prove : (hypotenuse)² = (Base)² + (perpendicular)²

construction : Drow BD | AC
proof : ∆ADB ~ ∆ABC.

[ if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse Then triangle on both sides of the perpendicular are similar to the whole triangle and to each other .]

so,
$\: \: \: \: \: \: \frac{AD}{AB} = \frac{AB}{AC} \\ \: \\ = > AD.AC \: = A {B}^{2}$
also, ∆BDC ~ ∆ABC

so,
$\: \: \: \: \: \: \: \frac{CD}{BC} = \frac{BC}{AC} \\ \\ = > CD.AC \: = B {C}^{2}$
Adding ( 1 ) and ( 2 ) , we have
$\: \: \: \: \: \: \: \: AD.AC \: + CD.AC \: = A {B}^{2} + B {C}^{2} \\ = > (AD + CD).AC \: = A {B}^{2} + B {C}^{2} \\ = > AC.AC \: = A {B}^{2} + B {C}^{2} \\ hence \: \: A {C}^{2} = A {B}^{2} + B {C}^{2}$
FOR THE SECOND PART:
we have
DB = 3CD (given)
$\: \: \: \: \: \: \: = > CD + DB = CD + 3CD \\ \: \: \: \: \: \: \: = > BC = 4CD$
in right angle ∆ABD , right angled at D , we have
AB² = AD² + BD²

=> 2AB² = 2AD² + 2BD² .......... (i)

In right angle ∆ABC , right angled at D , we have

AC² = AD² + DC²

=> 2AC² = 2AD² + 2DC². ..........(ii)

SUBTRACT (2) FROM (1) , WE GET
$2A {B}^{2} - 2A {C}^{2} = (2A {D}^{2} + 2B {D}^{2} ) - (2A {D}^{2} + 2D {C}^{2} ) \\ = > 2A {B }^{2} - 2A {C}^{2} = 2(A {B}^{2} - D {C}^{2} ) \\ = > 2A {B}^{2} - 2A {C}^{2} = 2(9D {C}^{2} - D {C}^{2} ) \\ = > 2A {B}^{2} - 2A {C}^{2} = 2(8D {C}^{2} ) \\ = > 2A {B}^{2} - 2A {C}^{2} = 16D {C}^{2} \\ = > 2A {B}^{2} - 2A {C}^{2} = (4D {C)}^{2} \\ = > 2A {B}^{2} - 2A {C}^{2} = (B {C)}^{2} \\ = > 2A {B}^{2} = 2A {C}^{2} + B {C}^{2}$

here is proved..

hope it's helps you .
☺☺