In mathematics, we say two things are **congruent** if they are the same, up to some kind of transformation.

For example, $\frac{1}{2} = \frac{2}{4}$ because they both represent the value $0.5$, though they have different numerators and denominators. In this sense, fractions are congruent if and only if their numerators and denominators are a constant multiple of each other.

In geometry, two figures are congruent if they have the exact same size and shape. The figures may be translated, rotated or even reflected. [Note: Some people distinguish a reflection by saying that it has a different orientation.] If two figures are congruent, they have the same corresponding lengths and angles, which might give us more information about the figures. Conversely, if we show that two figures have the same corresponding lengths *and* angles, then we know that they are congruent figures.

In the case of triangles, if two triangles have corresponding lengths (but we do not know anything about their angles), we can apply the Cosine Rule to conclude that they have the same corresponding angles. In fact, apart from SSS (side-side-side), we also have the methods of SAS (side-angle-side), ASA (angle-side-angle), and AAS (angle-angle-side) to prove triangles are congruent. Remember that SSA (side-side-angle) doesn't necessarily imply congruency, as in the Ambiguous case of Sine Rule.

## 1. In isosceles triangle $ABC$ with $AB=BC$, let $D$ be the foot of the perpendicular from $B$ to $AC$. Show that $ABD$ and $CBD$ are congruent triangles.

Solution: Since the triangle is isosceles, we know that $AB = BC$ and $\angle BAD = \angle BCD$. Then, $AD^2 = AB^2 - BD^2 = CB^2 - BD^2 = CD^2$ which gives us $AD = CD$. Hence, by SAS, we get that triangles $ABD$ and $CBD$ are congruent.

## 2. Prove that if two triangles have equal corresponding side lengths, then they are congruent.

Solution: Let $ABC$ and $DEF$ be triangles such that

$AB=DE, BC=EF, CA= FD.$

From the Cosine Rule, we know that $\cos \angle ABC = \frac{ AB^2 + BC^2 - CA^2}{2 AB \times BC } = \frac{ DE^2 + EF^2 - FD^2} { 2 DE \times EF} = \cos \angle DEF.$ Since angles in a triangle are in the range $0 ^ \circ$ to $180^\circ$, it follows that $\angle ABC = \angle DEF$. Similar equations hold for the other angles, which show that the corresponding angles are equal. Hence, the triangles are congruent.

## 3. $ABCD$ is a parallelogram. Let $AC$ and $BD$ intersect at $E$. Show that triangles $EAB$ and $ECD$ are congruent.

Solution: Since $ABCD$ is a parallelogram, it follows that $\angle EAB = \angle ECD$ and $\angle DBA = \angle EDC$. By vertical angles, $\angle AEB = \angle CED$ and by AAA (angle-angle-angle), the two triangles are similar.

To show the triangles are congruent, it remains to show that two corresponding sides of the triangle have the same length. This follows immediately since $ABCD$ is a parallelogram, implying $AB = CD$.

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