Crossed ladders theorem

Students will learn what the crossed ladders theorem is, and how to prove it. Then students will learn what the extended crossed ladders theorem is, and how to prove it. Both proofs can be found here. After that, students will learn how to prove the corollary

$$\dfrac{1}{\text{area}(\triangle AEC)} + \dfrac{1}{\text{area}(\triangle ADC)} = \dfrac{1}{\text{area}(\triangle AFC)} + \dfrac{1}{\text{area}(\triangle ABC)}$$

Here's the solution: We know the area formula for triangles is \((1/2)bh.\) We also know the base of each triangle is \(AC.\) To obtain the height, we can construct \(\overline{EI},\) \(\overline{DH},\) etc. to be perpendicular to \(\overline{AC}.\)

Next, we use the area formula for triangles, to obtain

$$\begin{align} & \text{area}(\triangle AEC) = \tfrac{1}{2}\overline{AC} \cdot \overline{EI} \\[0.5em] & \text{area}(\triangle ADC) = \tfrac{1}{2}\overline{AC} \cdot \overline{DH} \\[0.5em] & \text{area}(\triangle AFC) = \tfrac{1}{2}\overline{AC} \cdot \overline{FJ} \\[0.5em] & \text{area}(\triangle ABC) = \tfrac{1}{2}\overline{AC} \cdot \overline{BG} \end{align}$$

After that, we substitute these into our goal, then apply some basic algebra, giving us

$$\dfrac{1}{\overline{EI}} + \dfrac{1}{\overline{DH}} = \dfrac{1}{\overline{FJ}} + \dfrac{1}{\overline{BG}}$$

But this is true by the extended ladders theorem. Thus, we have proven our corollary.