Proofs with Triangles
We've just started our journey on constructing geometrical proofs using deductive reasoning.
Geometric proofs can be used in all sorts of ways:
- to show that various angles have particular values,
- to discover new angle relationships using known ones, or as we will practise in this lesson,
- to verify relationships within triangles.
When it comes to geometric proofs, there is no one formula or process to follow. It's a matter of using the information given in any which way possible to find angles and/or form relationships.
Example
Prove that $d+e=b+c$d+e=b+c 
Strategy:
Our strategy here is to form some relationships that link $b$b and $c$c to $d$d and $e$e. We can see that the angles belong to 2 triangles with $\angle DAE$∠DAE common to both triangles so let's try and use that:
Proof:
| Consider $\triangle ADE$△ADE | |||
| $\angle ADE+\angle DAE+\angle DEA$∠ADE+∠DAE+∠DEA | $=$= | $180$180 | (Angle sum of a triangle is $180$180°) |
| $\angle DAE+d+e$∠DAE+d+e | $=$= | $180$180 | Rearrange to make angle(DAE) the subject |
| $\angle DAE$∠DAE | $=$= | $180-\left(d+e\right)$180−(d+e) | |
| Consider $\triangle ABC$△ABC | |||
| $\angle CBA+\angle ACB+\angle BAC$∠CBA+∠ACB+∠BAC | $=$= | $180$180 | Angle sum of a triangle is $180$180° |
| $b+c+\angle BAC$b+c+∠BAC | $=$= | $180$180 | Rearrange to make angle(BAC) the subject |
| $\angle BAC$∠BAC | $=$= | $180-\left(b+c\right)$180−(b+c) | |
Now, notice that $\angle BAC$∠BAC and $\angle DAE$∠DAE are exactly the same angle.
This means we can equate the 2 expressions for $\angle BAC$∠BAC and $\angle DAE$∠DAE
| $180-\left(d+e\right)$180−(d+e) | $=$= | $180-\left(b+c\right)$180−(b+c) |
| $-\left(d+e\right)$−(d+e) | $=$= | $-\left(b+c\right)$−(b+c) |
| $d+e$d+e | $=$= | $c+b$c+b |
Let's have a look at some other proofs involving triangles.
Examples
Question 1
Question 2
In the diagram $AC$AC bisects $\angle BAD$∠BAD, and $DE=EF$DE=EF. By letting $\angle CAD=x$∠CAD=x , prove that $AC$AC is parallel to $DF$DF.
