# For a triangle is given the next relation for its surface 4A = ( a + b + c )( a + b - c) . Prove that the triangle is right-angled . a, b, c are sides of the triangle

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4A = (a+b+c)(a+b-c) by data.

Therefore, by Heron's formula

4sqrt{s(s-a)(s-b)(s-c)} = (a+b+c)(a+b-c0, where s = 2(a+b+c)

4sqrt(s(s-a)(s-b)s-c)) = 2s*2(s-c)

sqrt{s(s-a)(s-b0(s-c)} = s(s-c). Squaring we get:

s(s-a)(s-b)(s-c) = s^2(s-c)^2

(s-a)(s-b) = s(s-c)

(b+c-a)(a+c-b) = (a+b+c)(a+b-c)

c^2 - (a+b)^2 = (a+b)^2-c^2

c^2 -a^2-b^2 +2ab= (a^2+b^2+2ab -c^2

2c^2 = 2(a^2+b^2)

c^2 = a^2+b^2 . So Pythagorus theorem holds for the sides of the triangle and so the triangle is a right angled triangle.

sqrt{s(s-a)s-b)(s-c) = s(s-c) squaring,

s(s-a)(s-b)(s-c) = s^2 (s-c)^2. reducing by s(s-c)

(s-a)(s-b) = s(s-c) Or

s^2-s(a+b)+ab = s^2 -sc

-s(a+b+c) =

s

We’ll use the generalized Pythagorean theorem:

c^2=a^2+b^2-2*a*b* cos (angle between a and b)-a^2+b^2-c^2=2*a*b* cos (angle between a and b)

A=a*b*sin (angle between a and b)/2, where A is the aria of the triangle

From this relation, 2A= a*b*sin (angle between a and b)

4A=2* a*b*sin (angle between a and b)

Let’s analyze the relation from enunciation:

4A=(a+b+c)*(a+b-c)

We notice that if we group (a+b)=d, then

4A=(d+c)(d-c)

4A=d^2-c^2=(a+b)^2-c^2

4A= a^2+b^2+2a*b-c^2

Let’s substitute the formulas previously processed:

2* a*b*sin (angle between a and b)= 2*a*b* cos (angle between a and b) +2a*b

We notice that we could divide the expression by the product 2a*b

sin (angle between a and b)= cos (angle between a and b) +1

sin (angle between a and b)- cos (angle between a and b)=1

If we square raise:

[sin (angle between a and b)- cos (angle between a and b)]^2=1

Let’s substitute angle between a and b= t

(sin t – cos t)^2=1

(sin t)^2+ (cos t) ^2-2sin t cos t=1

But from the fundamental formula

(sin t)^2+ (cos t) ^2=1

1-2sin t cos t=0

2sin t cos t=0, so sin t=0 or cos t=0

An angle of a triangle can not have sin t =0, so we have to have cos t=o, so sin t= 90, so the triangle is rectangled.