Geometry

UK Secondary (7-11)

Proofs Using Similar Triangles

Lesson

Recall that if two triangles are similar, then one is an enlargement or reduction of the other.

Similar triangles have two very important features:

- All corresponding angles are equal.

- All pairs of corresponding sides are in the same ratio.

$\frac{A}{D}$`A``D`$=$=$\frac{B}{E}$`B``E`$=$=$\frac{C}{F}$`C``F`

There are three main methods we can use to prove that two triangles are similar.

**AAA (Angle, Angle, Angle)**- All corresponding angles are equal.

**SAS (Side, Angle, Side)**- Two pairs of corresponding sides are in the same ratio and the corresponding angles in between are equal.

**SSS (Side, Side, Side)**- All three pairs of corresponding sides are in the same ratio.

$\frac{A}{D}$`A``D`$=$=$\frac{B}{E}$`B``E`$=$=$\frac{C}{F}$`C``F`

You can revise these ideas here.

We're going to use our knowledge of similar triangles along with other geometric concepts to come up with some more general proofs.

Consider the following parallelogram.

**a)** Prove that $\triangle ABO\simeq\triangle DOC$△`A``B``O`⫻△`D``O``C`.

**b)** Hence, deduce that the diagonals of a parallelogram bisect each other.

**a)** We are going to prove that $\triangle ABO$△`A``B``O` and $\triangle DOC$△`D``O``C` are similar using an **AAA** proof.

Firstly, we know that $\angle AOB=\angle DOC$∠`A``O``B`=∠`D``O``C` because they are vertically opposite.

Secondly, we know that $AB\parallel DC$`A``B`∥`D``C`, because $ABCD$`A``B``C``D` is a parallelogram. Hence, $\angle OAB=\angle OCD$∠`O``A``B`=∠`O``C``D` because they are alternate angles.

Thirdly, $\angle ABO=\angle ODC$∠`A``B``O`=∠`O``D``C` for the same reason. They are alternate angles on parallel lines.

Hence, by our **AAA** proof, $\triangle ABO\simeq\triangle DOC$△`A``B``O`⫻△`D``O``C`.

**b)** We know that pairs of corresponding sides of similar triangles are all in the same ratio.

$\frac{AB}{DC}$`A``B``D``C`$=$=$\frac{BO}{DO}$`B``O``D``O`$=$=$\frac{AO}{CO}$`A``O``C``O`

But we also know that $ABCD$`A``B``C``D` is a parallelogram, and parallelograms have equal opposite sides. Hence, $AB=DC$`A``B`=`D``C`, which means that $\frac{AB}{DC}=1$`A``B``D``C`=1. Therefore, all pairs of corresponding sides are in the same ratio $1$1, in other words, equal.

So these triangles aren't just similar. They are congruent.

So, as we can see, the diagonals split each other into two equal halves. $AO=CO$`A``O`=`C``O` and $DO=BO$`D``O`=`B``O`. Hence, the diagonals of a parallelogram bisect each other.