Similarity,

In what follows, we’ll see that many—if not most—of our results shall rely on the proportionality of sides in **similar triangles**. A convenient statement is as follows.

Similarity, Given the similar triangles Δ*ABC* ∼ Δ*A’BC’*,

We have that,

Conversely, if

then triangles Δ*ABC* ∼ Δ*A’BC’* are similar.

## Proof

Note first that Δ*AA’C’* and Δ*CA’C’* clearly have the same areas, which implies that Δ*ABC’ *and Δ*CA’B* have the same area (being the previous common area plus the area of the common triangle Δ*A’BC’*).

Therefore,

In an entirely similar fashion, one can prove that

Conversely, assume that

In the figure below, point *C”* has been located so that the segment [*A’C”*] is parallel to [*AC*]. But then triangles Δ*ABC* and Δ*A’BC”* are similar,

and so,

i.e. that BC” = BC’ . This clearly implies that C’ = C”, and so [A’C’ ] is parallel to [AC]. From this, it immediately follows that triangles Δ ABC and ΔA’BC’ are similar.

## Exercises for Similarity

**01.** Let Δ*ABC* and Δ*A’B’C’* be given with *ABC* = A’B’C’ and

**02.**

In the figure to the right, *AD* = *rAB*, *AE* = *sAC*.

Show that

**03.** Let Δ*ABC* be a given triangle and let *Y, Z* be the midpoints of [*AC*], [*AB*], respectively. Show that (*XY *) is parallel with (*AB*). (This simple result is sometimes called the **Midpoint Theorem**)

**04.**

In Δ*ABC*, you are given that

where *x* is a positive real number. Assuming that the area of Δ*ABC* is 1, compute the area of Δ*XYZ* as a function of *x*.

**05.** Let *ABCD* be a quadrilateral and let *EF GH* be the quadrilateral formed by connecting the midpoints of the sides of *ABCD*. Prove that* EF GH* is a parallelogram.

**06.**

In the figure to the right, *ABCD* is a parallelogram, and *E* is a point on the segment [*AD*]. The point *F* is the intersection of lines (*BE*) and (*CD*). Prove that *AB* ×* FB* = *CF* × *BE*.

**07.**

In the figure to the correct, tangents to the circle at *B* and *C *meet at point *A*. *P* is found** **on the minor arc *BC* and the tangent to the circle at P meets the lines (*AB*) and (*AC*) at the points *D *and *E*, respectively. Prove that *DOE* = ½*BOC*, where *O* is that the** **center of the circle.

Related Lessons: Pythagorean theorem