Different teachers want different types of proofs. I will use "m(∠1)" to mean "the measure of angle 1", which is a number, therefore it can be divided by 2 and can be found equal to another number.
Assume that ∠A ≅ ∠X
with ray(AB) bisecting ∠A into the two smaller (inside) angles ∠1 and ∠2
and ray(XY) bisecting ∠X into the two smaller (inside) angles ∠3 and ∠4.
To show that ∠1 ≅ ∠3.
Since ∠A ≅ ∠X, m(∠A) = m(∠X). (If anlges are congruent, they must have the same measure.)
Multiplying both sides by ½: ½m(∠A) = ½m(∠X).
But, since ray(AB) bisects ∠A into ∠1 and ∠2, m(∠1) = ½m(∠A)
and, since ray(XY) bisects ∠X into ∠3 and ∠4, m(∠3) = ½m(∠X)
Since ½m(∠A) = ½m(∠X), m(∠1) = m(∠3).
Since m(∠1) = m(∠3), ∠1 ≅ ∠3.
If this doesn't make sense, post back.
