## Cotangent Identity

$$\begin{array}{cc} & |\overline{AC}|\,|\overline{FD}| = |AEDF| = |\overline{AB}|\,|\overline{ED}| \\[6pt] \to \qquad & \large{b^2\,\left(\,\cot A + \cot B\,\right) = c^2\,\left(\,\cot A + \cot C\,\right)} \\ \end{array}$$

Posted 23 September, 2015 by in Proof Without Words, Trigonography

## Half-Angles in a Triangle

My answer to this question at Math.StackExchange.com. $$\Large{c\;\sin\frac{A-B}{2} \;\equiv\; (a-b)\;\cos\frac{C}{2}}$$

Posted 12 August, 2015 by in Proof Without Words, Trigonography

## Product Identities

\begin{align} 2 \cos A \cos B &= \cos(A-B) \;+\; \cos(A+B) \\ 2 \sin A \;\sin B &= \cos(A-B) \;-\; \cos(A+B) \end{align}

Posted 25 February, 2015 by in Classroom, Proof Without Words, Trigonography, Uncategorized

## Proof Without Words: Nine-Point Circle Property

The “Nine-Point Circle” of a triangle $$\triangle ABC$$ contains the vertices of the medial triangle ($$\triangle RST$$) and orthic triangle ($$\triangle UVW$$), as well as points $$X$$, $$Y$$, $$Z$$ that bisect segments from $$\triangle ABC$$’s orthocenter ($$H$$) to its vertices. A question on Math.StackExchange.com asked for proof that the center of the nine-point circle […]