If the areas of two similar triangles are equal, prove that they are congruent.
First, let’s sketch two triangles to help us think through this proof. We’ll call the first triangle triangle 𝐴𝐵𝐶 and our second triangle triangle 𝐷𝐸𝐹. We’re given or we already know that the area of triangle 𝐴𝐵𝐶 is equal to the area of triangle 𝐷𝐸𝐹. We also know that triangle 𝐴𝐵𝐶 is similar to triangle 𝐷𝐸𝐹.
We can say that the ratio of side length 𝐴𝐵 to 𝐷𝐸 must be equal to the ratio of side length 𝐵𝐶 over 𝐸𝐹 is equal to the ratio of 𝐴𝐶 over 𝐷𝐹. We know that this is true because, in similar triangles, corresponding sides are all in the same proportion.
Something else we can say about similar triangles is the ratio of their area squared is equal to the square of the ratio of their sides. What does that mean? Area of triangle 𝐴𝐵𝐶 squared over the area of triangle 𝐷𝐸𝐹 squared is a ratio that’s equal to 𝐴𝐵 squared over 𝐷𝐸 squared, which is equal to 𝐵𝐶 squared over 𝐸𝐹 squared, which is equal to 𝐴𝐶 squared over 𝐷𝐹 squared. This statement again is true because these triangles are similar.
What we can say now is that the area of triangle 𝐴𝐵𝐶 squared over the area of triangle 𝐷𝐸𝐹 squared must be equal to one. It must be equal to one because we were already told that the areas are equal. And anything over itself equals one.
What we have now is the ratio of a side squared over its corresponding side squared must be equal to one. And we want to take these three side length ratios and set them all equal to one. 𝐴𝐵 squared over 𝐷𝐸 squared equals one. 𝐵𝐶 squared over 𝐸𝐹 squared equals one. And 𝐴𝐶 squared over 𝐷𝐹 squared equals one.
To get rid of this square ratio, we want to take the square root of both sides. On the left, we have 𝐴𝐵 over 𝐷𝐸. And on the right, we still have one; the square root of one is one. If the ratio of 𝐴𝐵 to 𝐷𝐸 equals one, then side length 𝐴𝐵 must be equal to side length 𝐷𝐸. That’s the only way they can have a ratio of one.
We’ve now proven that side length 𝐴𝐵 is equal in length to side length 𝐷𝐸. We follow the same procedure for our second ratio. Take the square root of both sides, and you have the ratio of 𝐵𝐶 to 𝐸𝐹 must be equal to one. Again, the only way for that to happen is for 𝐵𝐶 to be equal to 𝐸𝐹. This proves that side length 𝐵𝐶 is equal to side length 𝐸𝐹.
We do this one more time. Take the square roots of both sides. Our new statement says 𝐴𝐶 over 𝐷𝐹 must be equal to one. That means that 𝐴𝐶 must be equal to 𝐷𝐹. Side length 𝐴𝐶 is equal in length to 𝐷𝐹. We know that these two triangles must be congruent because all of their corresponding sides are equal, and proving congruent by side-side-side.