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...

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.