Question

if a:b=c:d then show that (a^2+b^2):(a^2-b^2)=(ac+bd):(ac-bd)

Answer 1

The other answers using algebra are the easiest way to get the answer.Just for fun, here's a method using vectors. If you know a bit about vectors is might help give an intuition as to why the result is true. The equation boils down to 1=sin2+cos2.Define vectors p and q in cartesian coordinates:p⃗ =(a,b,0)q⃗ =(c,d,0)and let θ be the angle between p⃗  and q⃗ .Then1=sin2θ+cos2θp2q2=p2q2sin2θ+p2q2cos2θ|p⃗ |2|q⃗ |2=(pqsinθ)2+(pqcosθ)2|p⃗ |2|q⃗ |2=(p⃗ ⋅q⃗ )2+|p⃗ ×q⃗ |2(a2+b2)(c2+d2)=(ac+bd)2+(ad−bc)2

Answer 2

Answer:

Let us start from the right side:

(ac + bd)^2 + (ad - bc)^2 = (ac)^2 + 2acbd + (bd)^2 + (ad)^2 - 2abcd + (bc)^2

Let us simplify:

(ax+bd)^2 + (ad-bc)^2 = (ac)^2 + (bd)^2 + (ad)^2 + (bc)^2

Let us rearrange terms:

==> (ac+ bd)^2 + (ad-bc)^2= (ac)^2 (bc)^2 + (bd)^2 + (ad)^2

Now we will factor:

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

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

==> (ac+bd)^2 + (ad-ac)^2 = (a^2 + b^2 )(c^2 + d^2)

Similarly:

(ac-bd)^2 + (ad+ac)^2 = (a^2 + b^2)((c^2 + d^2)