Trigonometry — The Angle Sum Theorem

Posted on 23 June, 2026

Definitions

A B C D

α=∠BADa=BC¯β1=∠ABDb1=AD¯β2=∠DBCb2=DC¯γ=∠BCDc=AB¯h=BD¯ \begin{align*} \alpha &= \angle BAD \\ a &= \overbar{BC} \\ \beta_1 &= \angle ABD \\ b_1 &= \overbar{AD} \\ \beta_2 &= \angle DBC \\ b_2 &= \overbar{DC} \\ \gamma &= \angle BCD \\ c &= \overbar{AB} \\ h &= \overbar{BD} \end{align*}

Observations

sin⁡α=cos⁡β1=hcsin⁡γ=cos⁡β2=hacos⁡α=sin⁡β1=b1ccos⁡γ=sin⁡β2=b2ac2=b12+h2a2=b22+h2 \begin{multline} \sin \alpha = \cos \beta_1 = \frac{h}{c} \\ \sin \gamma = \cos \beta_2 = \frac{h}{a} \\ \cos \alpha = \sin \beta_1 = \frac{b_1}{c} \\ \cos \gamma = \sin \beta_2 = \frac{b_2}{a} \\ c^2 = b_1^2 + h^2 \\ a^2 = b_2^2 + h^2 \end{multline}

Hypothesis

cos⁡(α+β1+β2+γ)=−1 \cos \left( \alpha + \beta_1 + \beta_2 + \gamma \right) = -1

Proof

cos⁡(α+β1+β2+γ)=−1⟹cos⁡(α+β1)cos⁡(β2+γ)−sin⁡(α+β1)sin⁡(β2+γ)=−1⟹(cosαcosβ1−sinαsinβ1)(cosβ2cosγ−sinβ2sinγ)−(sinαcosβ1+cosαsinβ1)(sinβ2cosγ+cosβ2sinγ)=−1⟹(b1c⋅hc−hc⋅b1c)(ha⋅b2a−b2a⋅ha)−(hc⋅hc+b1c⋅b1c)(b2a⋅b2a+ha⋅ha)=−1⟹0−(h2+b12c2)(b22+h2a2)=−1⟹0−(c2c2)(a2a2)=−1⟹0−1(1)=−1⟹0−1=−1∎ \begin{multline} \cos \left( \alpha + \beta_1 + \beta_2 + \gamma \right) = -1 \\ \implies \\ \cos \left( \alpha + \beta_1 \right) \cos \left( \beta_2 + \gamma \right) - \sin \left( \alpha + \beta_1 \right) \sin \left( \beta_2 + \gamma \right) = -1 \\ \implies \\ \left( \cos \alpha \cos \beta_1 - \sin \alpha \sin \beta_1 \right) \left( \cos \beta_2 \cos \gamma - \sin \beta_2 \sin \gamma \right) - \left( \sin \alpha \cos \beta_1 + \cos \alpha \sin \beta_1 \right) \left( \sin \beta_2 \cos \gamma + \cos \beta_2 \sin \gamma \right) = -1 \\ \implies \\ \left( \frac{b_1}{c} \cdot \frac{h}{c} - \frac{h}{c} \cdot \frac{b_1}{c} \right) \left( \frac{h}{a} \cdot \frac{b_2}{a} - \frac{b_2}{a} \cdot \frac{h}{a} \right) - \left( \frac{h}{c} \cdot \frac{h}{c} + \frac{b_1}{c} \cdot \frac{b_1}{c} \right) \left( \frac{b_2}{a} \cdot \frac{b_2}{a} + \frac{h}{a} \cdot \frac{h}{a} \right) = -1 \\ \implies \\ 0 - \left( \frac{h^2 + b_1^2}{c^2} \right) \left( \frac{b_2^2 + h^2}{a^2} \right) = -1 \\ \implies \\ 0 - \left( \frac{c^2}{c^2} \right) \left( \frac{a^2}{a^2} \right) = -1 \\ \implies \\ 0 - 1 \left( 1 \right) = -1 \\ \implies \\ 0 - 1 = -1 \\ \QED \end{multline}