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