Areas of Triangles and Parallelograms Proportional to Base

Theorem
Let $ABC$ and $ACD$ be triangles.

Let $EC, CF$ be parallelograms under the same height.

Then:
 * $AC : CD = \triangle ABC : \triangle ACD = \Box EC : \Box CF$

where:
 * $AC : CD$ denotes the ratio of the length of $AC$ to that of $CD$
 * $\triangle ABC : \triangle ACD$ denotes the ratio of the area of $\triangle ABC$ to that of $\triangle ACD$
 * $\Box EC : \Box CF$ denotes the ratio of the area of parallelogram $EC$ to that of parallelogram $CF$.

Proof

 * Euclid-VI-1.png

Let $BD$ be produced in both directions to the points $H, L$ and let any number of straight lines, for example, $BG, GH$ be made equal to the base $BC$, and any number of straight lines, for example, $DK, KL$ be made equal to the base $CD$.

Let $AG, AH, AK, AL$ be joined.

Since $CB = BG = GH$ it follows from Triangles with Equal Base and Same Height have Equal Area that $\triangle ABC = \triangle AGB = \triangle AHG$.

Therefore, whatever multiple the base $HC$ is of the base $BC$, that multiple also is $\triangle AHC$ of $\triangle ABC$.

For the same reason, whatever multiple the base $LC$ is of the base $CD$, that multiple also is $\triangle ALC$ of $\triangle ACD$.

If the base $HC$ is equal to the base $CL$, from Triangles with Equal Base and Same Height have Equal Area, $\triangle AHC = \triangle ACL$.

If the base $HC$ is greater than the base $CL$, $\triangle AHC > \triangle ACL$.

If the base $HC$ is less than the base $CL$, $\triangle AHC < \triangle ACL$.

So we have four magnitudes, two bases $BC, CD$ and two triangles $ABC, ACD$.

We also have that equimultiples have been taken of the base $BC$ and $\triangle ABC$, namely the base $HC$ and $\triangle AHC$.

Also we have equimultiples of the base $CD$ and $\triangle ADC$, namely the base $LC$ and $\triangle ALC$.

It has been proved that:
 * $HC > CL \implies \triangle AHC > ALC$
 * $HC = CL \implies \triangle AHC = ALC$
 * $HC < CL \implies \triangle AHC < ALC$

Therefore from $BC : CD = \triangle ABC : \triangle ACD$.

Next, from Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $EC$ is twice the area of $\triangle ABC$.

Similarly, the parallelogram $FC$ is twice the area of $\triangle ACD$.

So from Ratio Equals its Multiples, $\triangle ABC : \triangle ACD = \Box FC : \Box RC$.

The result follows from Equality of Ratios is Transitive.