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OpenStudy (dehydrated):
OpenStudy (dehydrated):
I KNOW THAT Alternate Interior Angles theorem states, if two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.
OpenStudy (dehydrated):
@Jaynator495
Directrix (directrix):
In the triangle congruence statement in the proof,
Point A corresponds to point C
So, the missing pair of alternate interior angles would be:
<CAB is congruent to <ACB
OpenStudy (dehydrated):
@Directrix
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OpenStudy (dehydrated):
TY SO MUCH
Directrix (directrix):
Hold on. Let me check. I may have mixed the letters.
OpenStudy (dehydrated):
ok
Directrix (directrix):
<DAC is congruent to <BCA
Now, to choose that option.
OpenStudy (dehydrated):
these are the only options @directrix
∠ABD ≅ ∠DBC
∠CAD ≅ ∠ACB
∠BDA ≅ ∠BDC
∠CAB ≅ ∠ACB
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Directrix (directrix):
I have drawn the figure and marked it up several times and cannot find an answer that equates to:
<BCA is congruent to <ADB.
I'll draw it out.