3.1 Indirect proof
In all of our proofs so far, we have started with the given information, often including a diagram, and have worked directly from step to step, having each new step follow as a direct and logical conclusion from what has proceeded. We now introduce a second form of proof. Rather than have each step of the proof lead directly to the next, we will get to our desired conclusion by the back door. If we make an assumption — which might be either true or false — and find that it leads to an impossibility or to a conclusion that contradicts a known fact, then the assumption must be false. Remember that a statement, in this case our assumption, must be either true or false.
Certain collections of statements are compatible with one another; others are not. For instance, the two statements “$X$ is a prime number” and “$X=6$” cannot both be true at the same time. If you begin a problem by assuming that $X$ is a prime number, and you discover after additional work that $X$ must equal 6, then the assumption that $X$ was prime must be false, since that assumption led you to an incompatible result. This is the basis of the reasoning in an indirect proof. We have a desired conclusion. We ask the question, “If that desired conclusion doesn’t happen, can that possibly be consistent with the given situation?” If not, then the desired conclusion must occur.
Let us look at an example of an indirect proof.
Example:
Given: In $\triangle ABC$,
$AB\neq AC$ and $\overline{AD}$ is a median. Prove: $\overline{AD}$ cannot be perpendicular to $\overline{BC}$. |
Proof: Either $\overline{AD}$ is perpendicular to $\overline{BC}$ or it is not. What happens if we assume that $\overline{AD}\perp\overline{BC}$ In that case, $\angle ADB$ and $\angle ADC$ would both be right angles and hence congruent. We are given that $\overline{AD}$ is a median, so $\overline{CD}\cong\overline{DB}$. Because $\overline{AD}\cong\overline{AD}$, it implies that $\triangle ADC\cong\triangle ADB$ by SAS. By CPCTC, we then have $\overline{AB}\cong\overline{AC}$; but this would contradict our given, that $AB\neq AC$. Therefore our assumption (that $\overline{AD}\perp\overline{BC}$) is false; $\overline{AD}$ cannot possibly be perpendicular to $\overline{BC}$.
Now we develop an idea that will help us with future theorems.
An exterior angle of a triangle is an angle formed by a side of a triangle and the extension of an adjacent side of the triangle. The remote interior angles are the angles of the triangle far away from the exterior angle.
The adjacent interior angle is supplementary to the exterior angle and it appears as if the exterior angle is greater than either of the remote interior angles. Let us now prove the Exterior Angle Inequality that verifies this.
- Theorem 3.1:
Exterior Angle Inequality
The measure of an exterior angle of a triangle is greater than the measure of either of its remote interior angles.
Given: $\angle BCD$
is an exterior angle of $\triangle ABC$. Prove: $m\angle BCD > m\angle B$ and $m\angle BCD > m\angle BAC$ |
Proof: Let $M$ be the midpoint of $\overline{BC}$. Extend $\overline{AM}$ to $P$ so that $MP=AM$. Draw $\overline{CP}$. By the definition of midpoint, $\overline{MB}\cong\overline{MC}$, also $\overline{AM}\cong\overline{MP}$, and the vertical angles $\angle AMB$ and $\angle PMC$ are congruent. Therefore $\triangle AMB\cong\triangle PMC$ and $\angle 1\cong \angle B$. Because $m\angle BCD > m\angle 1$, we know that $m\angle BCD > m\angle B$. We can prove the second part of the theorem by extending $\overline{BC}$ and mimicking what we have just done by using the median drawn from vertex $B$.
This theorem will be critical in proving some of the theorems that are coming up.