What is the name of this theorem and how do I prove it?

Question

A triangle is intersected by two transversals, both parallel to the base of the triangle. Two more triangles appear and all three are similar. Also the area of $cde=fdeg=fgab$.

Find the exact value of the quotient $$\frac{cd}{fa}$$

Answer 1

The three triangles are similar. So there are constants $r_1$ and $r_2$ so that $CF= r_1 CD$ and $CA = r_2 CD$ and $\frac {CD}{FA} = \frac {CD}{CA - CF} =\frac {C}{r_2 CD - r_1 CF} = \frac 1{r_2 - r_1}$.

The area of $\triangle CFG = A \triangle CDE + A( FDEG) = 2\times A \triangle CDE$ and the area of $\triangle CAB =A \triangle CDE + A(FGAB) = 3\times A \triangle CDE$

But if the base and height of $\triangle CDE$ are $b,h$ then the base and height of $\triangle CFG$ and $\triangle CAB$ are $r_1b,r_1h$ and $r_2b, r_2h$ and the area of the triangles are $A\triangle CFG = \frac 12r_1br_1h = (\frac 12bh)r_1^2 = r_1^2 A \triangle CDE=2A\triangle CDE$ and $A\triangle CAB = \frac 12r_2br_2h = (\frac 12bh)r_2^2 = r_2^2 A \triangle CDE=3A\triangle CDE$

So $r_1 = \sqrt 2$ and $r_2 = \sqrt 3$

So $\frac {CD}{FA} = \frac 1{r_2 - r_1} = \frac 1{\sqrt 3-\sqrt 2}$

That's all.

Answer 2

Since the smallest triangle and the two trapezoids have equal area, you know that

• the medium triangle has twice the area of the smallest triangle
• the largest triangle has three times the area of the smallest triangle.

Since the triangles are similar, this means the corresponding sides of the triangle are in a ratio that is the square root of the area ratio. For example, comparing the middle triangle with the smallest triangle, we have $\frac{\overline{CD}}{\overline{CF}} = \sqrt{\frac{1}{2}}$. This is how you relate the information about area ratios to reason about side lengths.