So....I need either an angle or a side.

[Wormman12]

I'm starting to get the hang of these, but this one has a bit of a confusion to it. from what I got from it, I figured this out:


Since M is the midpoint for line AC; Line AM is cong w/Line MC.
line BM perpendicualr to Line AC; ∠AMB≈∠CMB (both being right angles).

The next step, I want to say that line BM bisects B so I can have an ASA theorem to complete it, but that isn't given. I know that if we consider tat line AC is a transversal between lines BC and BA they are supplementary, but that doesn't mean they are congruent, so I am stumped.View attachment Mathmatical problem2.pdf

 

[Mrspi]

I'm starting to get the hang of these, but this one has a bit of a confusion to it. from what I got from it, I figured this out:


Since M is the midpoint for line AC; Line AM is cong w/Line MC.
line BM perpendicualr to Line AC; ∠AMB≈∠CMB (both being right angles).

The next step, I want to say that line BM bisects B so I can have an ASA theorem to complete it, but that isn't given. I know that if we consider tat line AC is a transversal between lines BC and BA they are supplementary, but that doesn't mean they are congruent, so I am stumped.View attachment 2568

Do you see that BM is a side of EACH of the triangles you are trying to prove congruent?

BM = BM by the reflexive property of equality. (or, you could say that BM is congruent to BM by the reflexive property of congruence.)

Either way, your two triangles are congruent by SAS.

 

[Wormman12]

Okay, so....whenever two angles use the same side, that side is considered congruent?

 

[Mrspi]

Okay, so....whenever two angles use the same side, that side is considered congruent?

If the same segment is a side of each of the two angles, then yes, that segment is surely congruent to itself. The reflexive property of congruence says that any segment is congruent to itself.