Areas of Triangles and Parallelograms Proportional to Base
Theorem
In the words of Euclid:
- Triangles and parallelograms which are under the same height are to one another as their bases.
(The Elements: Book $\text{VI}$: Proposition $1$)
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
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 Book $\text{V}$ Definition $5$: Equality of Ratios $BC : CD = \triangle ABC : \triangle ACD$.
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.
$\blacksquare$
Historical Note
This proof is Proposition $1$ of Book $\text{VI}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VI}$. Propositions