Answer: This statement is known as the Law of Sines and can be proven using trigonometry and the properties of similar triangles.
Consider a triangle ABC with sides of lengths a, b, and c opposite to the angles A, B, and C respectively. Let's draw an altitude from vertex A to side BC, splitting side BC into two segments: BD of length x and CD of length c-x.
Using the Pythagorean theorem, we can write:
a^2 = x^2 + h^2 (1)
b^2 = (c-x)^2 + h^2 (2)
where h is the length of the altitude from A to BC.
Dividing equation (1) by sin^2(A) and equation (2) by sin^2(B), we get:
a^2 / sin^2(A) = x^2 / sin^2(A) + h^2 / sin^2(A)
b^2 / sin^2(B) = (c-x)^2 / sin^2(B) + h^2 / sin^2(B)
Since angles A and B are complementary (i.e. A + B = 90 degrees), we have sin(A) = cos(B) and sin(B) = cos(A). Substituting these identities and rearranging the equations, we get:
a^2 / sin(A) = x / cos(B) + h / sin(B)
b^2 / sin(B) = (c-x) / cos(A) + h / sin(A)
Multiplying both equations by sin(A)sin(B), we obtain:
a^2 sin(B) = x sin(A) cos(B) + h sin(A)
b^2 sin(A) = (c-x) sin(B) cos(A) + h sin(B)
Now, we use the fact that h = a sin(B) = b sin(A), which follows from the definition of sine as opposite/hypotenuse. Substituting this into the above equations and simplifying, we get:
a / sin(A) = 2R
b / sin(B) = 2R
c / sin(C) = 2R
where R = a/(2sin(A)) = b/(2sin(B)) = c/(2sin(C)) is the radius of the circumcircle of triangle ABC. This is the Law of Sines in its usual form.
From this, we can see that the ratio of the sines of any two angles in a triangle equals the ratio of the lengths of their opposite sides, as required.
Explanation: