Question

A, B, C, D are four points taken on a line in that order. AB = a, BC = b, CD = c. P is a point outside this line. Measures of angles APB = BPC = CPD = θ. Prove that cos^2(θ) = (a + b)(b + c)/4ac

Answer 1

Answer on Question #48117 – Math – Geometry

A, B, C, D are four points taken on a line in that order. AB = a, BC = b, CD = c. P is a point outside this line. Measures of angles APB = BPC = CPD = θ. Prove that $\cos^2(\theta) = (a + b)(b + c)/4ac$.

Solution.

Denote $AP = x, BP = y, CP = z, DP = t$.

By the bisector theorem we have:

$$\frac{AB}{AP} = \frac{BC}{CP} \Rightarrow \frac{a}{x} = \frac{b}{z} \Rightarrow z = \frac{b}{a} x;$$ $$\frac{BC}{BP} = \frac{CD}{PD} \Rightarrow \frac{b}{y} = \frac{c}{t} \Rightarrow t = \frac{c}{b} y;$$ $$S_{APC} = S_{APB} + S_{BPC} \Rightarrow xz \frac{\sin 2\theta}{2} = xy \frac{\sin \theta}{2} + yz \frac{\sin \theta}{2} \Rightarrow$$ $$\Rightarrow xy + yz = 2xz \cos \theta \Rightarrow y = \frac{2xz \cos \theta}{x + z} = \frac{2x \frac{b}{a} x \cos \theta}{x + \frac{b}{a} x} = \frac{2bx \cos \theta}{a + b};$$

Hence:

$$t = \frac{c}{b} y = \frac{2cx \cos \theta}{a + b};$$ $$S_{APD} = S_{APB} + S_{BPC} + S_{CPD} \Rightarrow \frac{xt \sin 3\theta}{2} = \frac{xy \sin \theta}{2} + \frac{yz \sin \theta}{2} + \frac{zt \sin \theta}{2} \Rightarrow$$ $$\Rightarrow xt(3 \sin \theta - 4 \sin^3 \theta) = \sin \theta (xy + yz + zt) \Rightarrow$$ $$\begin{array}{l} \Rightarrow xt(3 - 4 \sin^2 \theta) = xy + yz + zt \Rightarrow 4 \cos^2 \theta - 1 = \frac{xy + yz + zt}{xt} \Rightarrow \\ \Rightarrow \cos^2 \theta = \frac{xy + yz + zt + xt}{4xt} = \frac{(x + z)(y + t)}{4xt} = \frac{\left(x + \frac{b}{a}x\right)\left(y + \frac{c}{b}y\right)}{4x \cdot \frac{2cx \cos \theta}{a + b}} = \\ = \frac{\left(1 + \frac{b}{a}\right)\left(1 + \frac{c}{b}\right) \frac{2bx \cos \theta}{a + b}}{\frac{8cx \cos \theta}{a + b}} = \frac{\left(1 + \frac{b}{a}\right)\left(1 + \frac{c}{b}\right)b}{4c} = \frac{(a + b)(b + c)}{4ac}. \end{array}$$

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