Question

Fred and Victoria provide the following proofs for vertical angles to be equal.

Answer 1

Fred's proof: angle 2 + angle 3 = 180° (t is a straight line) angle 1 + angle 2 = 180° (PQ is a straight line) Therefore, angle 1 + angle 2 = angle 2 + angle 3 (Transitive Property of Equality) Hence, angle 1 = angle 3 (Subtraction Property of Equality)





Victoria's proof: angle 1 + angle 4 = 180° (t is a straight line) angle 1 + angle 2 =180° (PQ is a straight line) Therefore, angle 1 + angle 2 = angle 1 + angle 4 (Transitive Property of Equality) Hence, angle 2 = angle 4(Subtraction Property of Equality

Answer 2

Which statement is correct?

Both Fred's and Victoria's proofs are correct.

Both Fred's and Victoria's proofs are incorrect.

Only Fred's proof is correct.

Only Victoria's proof is correct.

Answer 3

i think that both fred and victoria are correct, but i wanna be sure and an explanation would be nice.

Answer 4

It is correct, they both applied the same technique, and the proceidure to prove the same thing, with the bare difference that they chose different angles to analyze, in the end proving two different equalities on the same scenario.
You see, a proof is a series of steps taken, proposition lead to another in a logical fashion leading thus to the Thesis, we start with given informaton called "Hypothesis" from which we conclude our first preposition:
\[H \rightarrow p_1\]
\[p_1\rightarrow p_2\]
.
.
.
\[p _{n-1} \rightarrow p_n\]
\[p_n \rightarrow T\]
This is pretty much a model for any mathematical proof, the hypothesis in this case is the intersection of two lines that generate four angles, and the thesis being vertical angles are equal.
I'll give you an example by provinf the identity: \((x+a)^2 = x^2+2xa+a^2\)
We will start with our hypothesis: \[(x+a)^2\]
And to reach preposition one, we will use the definition of exponent:
\[(x+a)^2 \rightarrow (x+a)(x+a)\]
And using the distributive property axiom we reach:
\[(x+a)(x+a) \rightarrow x^2 +xa +ax+ a^2 \]
Using the conmutitative property axiom we can conclude another preposition:
\[xa=ax \rightarrow xa+ax=2xa\]
Therefore, reaching our thesis:
\[x^2+2xa+a^2\]
We could've changed it a bit, and instead of using the definition of exponent, we would divide and multiply by (x+a) to reach the first proposition:
\[(x+a)^2 \rightarrow \frac{ (x+a)(x+a)^2 }{ (x+a) } \rightarrow (x+a)(x+a)\]
And the proof would've just go on as the last one, but you see, the information used was different, the conclusion was the same.

Answer 5

thank you so much! i skimmed over your answer but im reading it over thoroughly right now!~ thank you! ^.^