Similar Triangles

Two triangles that have corresponding congruent angles where the ratio of corresponding sides are equal are similar triangles.

Case 1:

Right Triangles Leg to Leg Case

Given the two triangles below, let's form a ratio of corresponding legs for the two right triangles.

Notice that if the ratio of the corresponding sides are equal, then the triangles are similar.

so, does ay=bx?

Let's try to prove this using the picture below.

Notice that Triangle IGH is congruent to triangle AKB because they are the same triangles.

Triangle JHK is congruent to itself.

Triangle IGH + Triangle JHK plus the rectangle with area bx form a larger triangle FGK.

Just as well, triangle AKB + Triangle JHK plus the rectangle with area ay form a larger triangle EHB.

So, Triangle FGK is congruent to triangle EHB by Angle-Side-Angle Congruence as well as Side-Angle-Side Congruence. Therefore,

since triangle FGK is congruent to triangle EHB.

By cancelling out the triangles that are congruent, we are left with bx = ay. Therefore the triangles are similar because the corresponding sides are equal. This proves the leg to leg case for right triangles.

Case 2:

Right Triangles Hypotenuse to Leg Case


a and x are legs of the right triangles and c and z are hypotenuses of the two right triangles.

So, does (leg)(hypotenuse) = (leg)(hypotenuse)?

Let's try using another picture to help us out.


We have the triangles we started with and we pull the larger ones c units apart from each other. The base of the parallelogram is z that is formed.

We can find the area of the whole rectangle in two ways.

First: The rectangle's base times its height is equal to

(y+c)x = xy + cx. Since xy is the area of the two large triangles, cx must represent the area of the parallelogram in between. This area can also be stated as az.

So (y+c)x also equals xy + az.

Therefore, xy + cx = xy + az.

By cancelling out the xy, we are left with cx = az.

This is what we wanted to prove so we have proved that the triangles are similar since the corresponding sides are in a fixed ratio by the leg to hypotenuse case for right triangles.

Theorem: Given 2 triangles, if the corresponding angles are congruent, then the triangles are similar.

Let's try to prove this!



Let h and H be the heights of the triangles in the picture below.


Let n = H/h = A/a = C/c

These arethe ratios of the lengths of the sides. Since they represent lengths, their ratios will give you a number.

so B1 = b1n and B2 = b2n.

Since B = B1 + B2, B = b1n + b2n.

Factoring out an n gives you B = n(b1 + b2).

Just as well b1 + b2 = b so B = n(b)

or B/b = n.