Question

open challenge to all !!!!!

Answer 1

aloha!

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here is your answer:

to prove:
${c}^{2} = {a}^{2} + {b}^{2} + 4(ar \: abc \: )$

construction: draw AK perpendicular to BC such that ABK is a right angled triangle.
where angle K is equal to 90°

proof: angle ACK = 180° - 135° = 45° { straight angle }

now,

tan C =
$\frac{ak}{ck}$
tan 45° =
$\frac{ak}{ck}$
1 =
$\frac{ak}{ck}$

AK = CK ----------------------- (1)

now,

c^2 = Ak^2 + BK^2
c^2 = AK^2 + ( a + CK )^2
c^2 = Ak^2 + a^2 + CK^2 + 2 × a × CK
c^2 = AK^2 + CK^2 + a^2 + 2 × a × CK
c^2 = b^2 + a^2 + 4 ( 1/2 × BC × AK )
c^2 = b^2 + a^2 + 4 ( ar ABC )


hence proved.


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hope it helps ;^)