Proving an angle within a triangle is congruent to another?

Question

It's been a while since I've done conventional geometry. How can I verify that angle $A$ is congruent to angle $C$?

I have no idea how to go about this without more info. Thanks.

Answer 1

The angles $A$ and $B$ are complementary. So are the angles $B$ and $C$. Therefore, the angles $A$ and $C$ are equal.

Answer 2

Hint: $$A+B=90=B+C.$$ Can you figure out?

Answer 3

Observe that $A$ and $B$ add up to $90$ (since $A+B+90=180$)

Hence,

$$ B=90-A $$

But, there is a triangle with angles $B$, $C$ and $90$. Hence by the angle sum property,

$$ B+C+90=180 $$

or,

$$ 90-A+C+90=180 $$

or,

$$ C-A+180=180 $$

or

$$ C-A=0 $$

finally!,

$$ C=A $$

QED

Some links which you might find useful to get back into math shape,

Triangles - http://artofproblemsolving.com/wiki/index.php?title=Triangle#Related_Formulae_and_Theorems

Similarity - http://artofproblemsolving.com/wiki/index.php?title=Similarity

Congruence - http://artofproblemsolving.com/wiki/index.php?title=Congruent

Short Hack for a very quick answer

In such problems usually (with lots of practice), you should be quickly able to guess that the answer is either $C=A$ or $C=90-A$.

After you are able to get this sort of intuition, the problem is very very easy to solve (especially if it is a MCQ question where no solution is required)

Re-draw the diagram by making the angle $A$ very small (almost zero), then check if angle $C$ is very small (almost zero) or big (almost 90)

If $C$ is small then the answer is $C=A$

If $C$ is big and almost equal to $90$ then the answer is $C=90-A$

Try this method yourself!