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First you have to prove the identity b^2=a^2+ac for such triangles. It can be proved using the similar triangles in the diagram or alternatively from the Sine and Cosine Rules.
Method 1
By construction \Delta ABX is an isosceles triangle. The angle of
this triangle at B is 180^\circ - 2\alpha, and hence the angles
at A and X are \alpha. Therefore \Delta XAC and \Delta
ABC are similar; hence
{a\over b} = {BC\over AC} = {AC\over XC}={b\over a+c},
u | v | a | b | c | |||
2 | 3 | 4 | 6 | 5 | |||
3 | 4 | 9 | 12 | 7 | |||
3 | 5 | 9 | 15 | 16 | |||
4 | 5 | 16 | 20 | 9 |
Note: it can be proved that a, b and c give a triangle if and only if v > u > v/2; see Math. Gazette , Vol. 412, June 1976, p.130.
Two trees 20 metres and 30 metres long, lean across a passageway between two vertical walls. They cross at a point 8 metres above the ground. What is the distance between the foot of the trees?
Three triangles ABC, CBD and ABD (where D is a point on AC) are all isosceles. Find all the angles. Prove that the ratio of AB to BC is equal to the golden ratio.
ABCD is a rectangle and P, Q, R and S are moveable points on the edges dividing the edges in certain ratios. Strangely PQRS is always a cyclic quadrilateral and you can find the angles.