Equivalence Criterion for Triangles
Two triangles are equivalent when they have congruent bases and corresponding heights.
In other words, when two triangles share the same base and height, they have the same area (equivalent surface).
Therefore, they are equivalent polygons.
A Practical Example
Let's consider two non-congruent triangles as an example.
Both triangles have the same base, b=3, and the same height, h=6.
The area of a triangle is calculated as half the product of its base and height.
$$ Area = \frac{b \cdot h}{2} $$
Since b=3 and h=6 for both triangles, their areas are the same:
$$ Area = \frac{3 \cdot 6 }{2} = \frac{18}{2} = 9 $$
Thus, with an area of 9, the two triangles are equivalent polygons.
The Proof
Consider two non-congruent triangles ABC and DEF.
By hypothesis, the two triangles have the same base and height.
$$ \overline{AB} \cong \overline{DE} $$
$$ \overline{CH} \cong \overline{DK} $$
We need to prove that they are equivalent polygons, meaning they have the same area.
Given that the area of a triangle is calculated as the product of its base and height divided by two,
$$ Area (triangle) = \frac{base \cdot height}{2} $$
Since triangles ABC and DEF have the same base AB≅DE and the same height CH≅FK, their areas are equal:
$$ Area (ABC) = Area(DEF) = \frac{\overline{AB} \cdot \overline{CH}}{2} = \frac{\overline{DE} \cdot \overline{FK}}{2} $$
Thus, the two triangles are equivalent.
This is sufficient to prove the initial theorem.
Alternative Proof
To prove that triangles ABC and DEF are equivalent, we can use another method.
Draw lines parallel to the bases AB and DE through the opposite vertices C and F.
Next, draw lines parallel to the sides BC and DF through the opposite vertices C and F.
This results in two parallelograms: ABCG and DEFJ.
The parallelograms ABCG and DEFJ each have twice the area of the triangles ABC and DEF, respectively.
$$ Area(ABCG) = 2 \cdot Area(ABC) $$
$$ Area(DEFJ) = 2 \cdot Area(DEF) $$
According to the parallelogram equivalence theorem, parallelograms ABCG and DEFJ are equivalent because they have congruent bases AB≅DE and congruent heights CH≅FK.
Therefore, they have equivalent areas:
$$ ABCG \doteq DEFJ $$
which means the same area:
$$ Area(ABCG) = Area(DEFJ) $$
Consequently, the area of parallelogram ABCG is twice that of triangle ABC and also twice that of triangle DEF.
$$ Area(ABCG) = 2 \cdot Area(ABC) = 2 \cdot Area(DEF) = Area(DEFJ) $$
Dividing everything by two:
$$ \frac{Area(ABCG)}{2} = \frac{2 \cdot Area(ABC)}{2} = \frac{2 \cdot Area(DEF)}{2} = \frac{Area(DEFJ)}{2} $$
$$ \frac{Area(ABCG)}{2} = Area(ABC) = Area(DEF) = \frac{Area(DEFJ)}{2} $$
From this, we deduce that the areas of triangles ABC and DEF are equal:
$$ Area(ABC) = Area(DEF) $$
This means that the two triangles ABC and DEF have equivalent areas.
Therefore, the two triangles are equivalent.
$$ Area(ABC) \doteq Area(DEF) $$
This concludes the proof.