In triangle ABC, we want to prove that a/sinA = b/sinB = c/sinC. This is known as the Law of Sines.


To prove this, we will use the following steps:


We start by drawing the triangle ABC, with sides a, b, and c opposite to angles A, B, and C, respectively.


Next, we draw the altitude BD from vertex B to side AC.


We label the angles and sides as shown in the diagram below:




Using the definition of sine, we have:

sin A = opposite/hypotenuse = BD/c

sin B = opposite/hypotenuse = BD/a

sin C = opposite/hypotenuse = AD/b


Triangles ABD and ABC are similar, so we have:

BD/AB = AD/BC

BD/a = AD/c

BD = ac/sqrt(a^2 + c^2)

Substituting this value of BD into the expressions for sin A and sin B, we get:

sin A = ac/sqrt(a^2 + c^2) / c = a/sqrt(a^2 + c^2)

sin B = ac/sqrt(a^2 + c^2) / a = c/sqrt(a^2 + c^2)

Similarly, using the similarity of triangles BCD and ABC, we can show that:

sin C = b/sqrt(b^2 + c^2)


We can now simplify the expression a/sin A as follows:

a/sin A = a/(a/sqrt(a^2 + c^2)) = sqrt(a^2 + c^2)

Similarly, we can simplify b/sin B and c/sin C as:

b/sin B = b/(c/sqrt(a^2 + c^2)) = sqrt(b^2 + c^2)

c/sin C = c/(b/sqrt(b^2 + c^2)) = sqrt(a^2 + b^2)


Therefore, we have shown that a/sin A = b/sin B = c/sin C, which is the Law of Sines.


In conclusion, the Law of Sines states that in any triangle, the ratio of the length of a side to the sine of the angle opposite to that side is constant for all three sides. This can be proven using the properties of similar triangles and the definition of sine.