**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 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__

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

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

**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.