Name the theorem or postulate that lets you conclude triangle abd is congruent to triangle cbd with only one additional step

The answer is the first one ⇒ AAS

Step-by-step explanation:

∵ DB ⊥ AC

∴ m∠DBA = m∠DBC = 90° ⇒ (1)

∵ m∠A = m∠C ⇒ (2) given (from figure)

∵ DB is a common side in the two Δ ⇒ (3)

From (1) , (2) and (3)

∴ Δ ABD is congruent to Δ CBD ⇒ AAS

answer: i'm great at algebra

~coco

step-by-step explanation:

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