# How would you break geometry proofs down into the necessary steps?

Nov 21, 2015

There are a few methodologies (see below), but the three most important ones are practice, practice and practice.

#### Explanation:

Proof is a set of logical steps that lead you from the given preposition statements to a statement to be proven.
In a way, it is like finding a way in a labyrinth from the starting point (given) to the ending point (to be proven).

One of the most universal methodologies of finding these steps is analysis-based approach.
This method starts from the ending point and determines which step might be leading to it. Finding one, we find another step that leads to the step we found. Unraveling this backwards step by step we have to get to the beginning statement given as the premises.

After this process (analysis) is completed, we should be able to prove our statement by going from the beginning reversing each step as we go until we reach a conclusion.

For example, let's prove that in an isosceles triangle $\Delta A B C$ ($A B \cong A C$) two medians, $B M$ and $C N$ to congruent sides are congruent.

Analysis

(Step 1) In order to prove congruence of two segments, we can include them into congruent triangles.

(Step 2) Let's choose triangles $\Delta A B M$ and $\Delta A C N$. Proving their congruence will suffice to state that $B M \cong C N$.

(Step 3) In order to prove congruence of triangles $\Delta A B M$ and $\Delta A C N$ we can find three pairs of congruent elements of these triangles.
3.1. They have a common angle $\angle B A C$
3.2. Congruent sides $A B$ in $\Delta A B M$ and $A C$ in $\Delta A C N$ (since both are legs of isosceles triangle $\Delta A B C$)
3.3. Congruent sides $A M$ in $\Delta A B M$ and $A N$ in $\Delta A C N$ (since they all are equal to half of a side $A B$).

This completes our analysis. But this is not the proof yet.
The real proof is the way backwards.

Proof
(reverse the logic from the given preposition to a statement to be proven)

1. Since $A B = A C$,
=> $A M = \frac{A C}{2} = \frac{A B}{2} = A N$

2. Since $A B = A C$ and $A M = A N$ and $\angle B A C$ is common for triangles $\Delta A B M$ and $\Delta A C N$,
=> $\Delta A B M$ ~= $\Delta A C N$

3. Since $\Delta A B M$ ~= $\Delta A C N$,
=> $B M = C N$

End of proof.