A = 36^o, AB = AC, C-bisector intersects AB at D, find BC without trig

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A = 36^o, AB = AC, C-bisector intersects AB at D, find BC without trig

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Source: Spanish Mathematical Olympiad 1994 P4

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#1 h Jul 31, 2018, 11:04 AM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ with $\angle A = 36^o$ and $AB = AC$ , the bisector of the angle at $C$ meets the oposite side at $D$ . Compute the angles of $\triangle BCD$ . Express the length of side $BC$ in terms of the length $b$ of side $AC$ without using trigonometric functions.

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#2 h Jul 31, 2018, 11:46 AM • 2 Y

Y by Adventure10, Mango247

Are you sure this was on SMO? Seems way to easy.

Anyway here's the solution.
With elementary angle chasing we get that BDC=72°, DCB=36° and DBC=72°. From angles, AD=CD=CB. As AC=AB=> DB= b-AD
Let's say AD=CD=CB=x

However, ∆ABC~∆DBC so x²=b²-bx. Usung quadratic formula we get x=(-b+b√5)/2.
As sides are CB=x=CD=b(√5-1)/2 and DB=b(3-√5)/2

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#3 h Aug 2, 2018, 7:58 AM • 1 Y

Y by Adventure10

$BC$ is the sidelength of regular decagon inscribed in a circle of radius $b.$ Therefore, $\boxed{BC=\frac{b}{2}(\sqrt 5-1)}$

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