Proving with congruent triangles

A proof in geometry is a chain of claims, and every claim has a reason: a given, a theorem we have learned, or earlier claims of the proof.

• The proof is written in a table: the claim in one column, the reason in the other.
• The rows are numbered. A claim that rests on earlier rows gives their numbers, such as “(1), (2), (3)”.

Once $\displaystyle \triangle ABC\cong\triangle DEF$ is proved, every pair of their corresponding parts is equal – and this is how we prove that segments or angles are equal.

• The correspondence is read from the order of the letters in the congruence.
• The reason for equal sides: “in congruent triangles, equal angles face equal sides”.
• The reason for equal angles: “in congruent triangles, equal sides face equal angles”.

This is how a proof with congruent triangles is built:

1. Mark all the givens in the drawing.
2. Look for two triangles that have the segments (or angles) to be proved equal as their sides (or angles).
3. Find three pairs of equal parts in them: givens, a common side, a common angle, vertical angles, alternate or corresponding angles between parallel lines.
4. Write the congruence by a congruence theorem, and then the conclusion.

In the quadrilateral $\displaystyle ABCD$: $\displaystyle {AB}={AD}$.
The diagonal $\displaystyle AC$ bisects the angle $\displaystyle BAD$: $\displaystyle {\angle BAC}={\angle DAC}$.
Prove: $\displaystyle {BC}={DC}$, and $\displaystyle {\angle ABC}={\angle ADC}$.

The proof below shows that triangles $\displaystyle ABC$ and $\displaystyle ADC$ are congruent – the side $\displaystyle AC$ is common to both – and draws both conclusions from the congruence:

The segments $\displaystyle AE$ and $\displaystyle BD$ cross at the point $\displaystyle C$.
$\displaystyle AB\parallel DE$.
$\displaystyle C$ is the midpoint of $\displaystyle BD$.
Prove: $\displaystyle {AB}={DE}$.

The proof below shows that triangles $\displaystyle ABC$ and $\displaystyle EDC$ are congruent: the angles at $\displaystyle B$ and $\displaystyle D$ are alternate angles between the parallel lines, and the angles at $\displaystyle C$ are vertical angles.

The segments $\displaystyle AC$ and $\displaystyle BD$ cross at the point $\displaystyle O$.
$\displaystyle O$ is the midpoint of $\displaystyle AC$, and $\displaystyle O$ is the midpoint of $\displaystyle BD$.
Prove: $\displaystyle {AB}={CD}$.

In triangle $\displaystyle ABC$: $\displaystyle {AB}={AC}$.
The point $\displaystyle D$ is on $\displaystyle AB$, and the point $\displaystyle E$ is on $\displaystyle AC$, so that $\displaystyle {AD}={AE}$.
Prove: $\displaystyle {BE}={CD}$.

In the quadrilateral $\displaystyle ABCD$: $\displaystyle {\angle B}={\angle D}={90^{\circ}}$.
$\displaystyle {AB}={AD}$.
Prove: $\displaystyle {BC}={DC}$.