So I was studying trigonometric ratios for right angled triangle, and I tried to prove that the trigonometric ratios for an angle \(\theta \) are constant for all right triangles with an angle \(\theta \). It is obvious that we can prove this with the help of similarity of triangles. This constantness of trigonometric ratios made me wonder that we can have constant ratios for Isosceles triangles too!

Assume we have an Isosceles triangle \(\vartriangle ABC\) with \(AB = AC = a\) and let say the included angle(between the equal sides) be \(\theta \). Let the length of the base be \(d\).

By similarity of triangles, all isosceles triangles with included angle(between the sides) equal to \(\theta \) are similar by AAA or AA rule.

Assume a similar isosceles triangle \(\vartriangle PQR\) with included angle \(\theta, PQ=PR=m \).
By the properties of similar triangles, we have: \(\tfrac{a}{m} = \tfrac{d}{QR}\), or \(\tfrac{d}{a} = \tfrac{QR}{m}\). And there we have it, the ratio of base : leg is constant for all isosceles triangles with included angle = \(\theta\).

NOTE that if we apply the cosine law on our \(\vartriangle ABC\), we get a relation between \(d,a\) such that:
\(\tfrac{d}{a} = 2sin(\tfrac{\theta}{2})\).

So now, let us invent a new ratio \(sl(\theta)\)
and define it as \(sl(\theta) = \tfrac{base(d)}{leg(a)} = 2sin(\tfrac{\theta}{2})\)

So basically just like \(sin(\theta)\) which is the ratio of opposite side to the hypotenuse for a right triangle, similarly we have \(sl(\theta)\) is the ratio of base to the leg and is equal to \(2sin(\tfrac{\theta}{2})\) in an ISOSCELES TRIANGLE