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**e^(i*pi)** It's an equilateral triangle so all sides must be the same length, it is one of the basic rules of geometry.

$\displaystyle AB = AY+YB = 2+1 = 3$ so you get $\displaystyle \dfrac{AY}{AB} = \dfrac{2}{3}$

I don't get how you're using the product rule. Can you explain how you're using line BC to get line YZ? Surely there is another line you're using.

**I was reading up on similar triangles and thought I found a good example, however after the event, "as always" I found that the similar traingles I was using were not equilateral, where after working through the maths I found that the sides in that example AC and AB were not equal, therefore a different method had been used, which was no good for my example.**

**edit: If I'm understanding correctly you've done $\displaystyle BC \times \dfrac{AY}{AB}$ using your carried forward error in $\displaystyle \dfrac{AY}{AB}$.**

**If you do $\displaystyle YZ = BC \times \dfrac{AY}{AB} = 3 \times \dfrac{2}{3} = 2$**

True, hence my confusion earlier

I shall use the cos rule to show YZ = 2.

AZ = AY = 2 - we are told this. We also know angle A is 60 degrees as it's a property of equilateral triangles.

$\displaystyle c^2 = a^2+b^2 - 2ab\cos(C)$. In this example **c = YZ** - the side we want to know, **a = AZ = 2**, **b = AY = 2** and **C = A = 60 degrees** (angle A)

$\displaystyle c^2 = 2^2 + 2^2 - 2(2)(2)\cos(60) = 4+4-4 = 4$ and so $\displaystyle c = YZ =2$

Thus YZ is 2 units across. It has been established that BC = 3 and so the ratio of YZ to BC is 2:3

Similarly the ratio of BC to YZ is the reciprocal which is 3:2