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

## Archive for the ‘**Proof Without Words**’ Category

## Cotangent Identity

## Half-Angles in a Triangle

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

## Product Identities

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

## 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.comĀ asked for proof that the center of the nine-point circle […]