AD = BC
AD || BC
According to which theorem are the triangles ΔADB≅ΔCBD congruent?
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AD = BC
AD || BC
According to which theorem are the triangles ΔADB≅ΔCBD congruent?
We are given: AD=BC
The angle ADB is equal to the angle CBD since AD is parallel to BC and the corresponding angles are equal between parallel lines.
DB=DB since it is a common side.
Therefore, we have two triangles that are congruent according to the S.A.S. (side, angle, side) theorem.
According to the S.A.S. theorem
Look at the triangles in the diagram.
Which of the statements is true?
When two lines are parallel and cut by a transversal, corresponding angles are equal. In this problem, DB acts as the transversal cutting parallel lines AD and BC, making ∠ADB = ∠CBD.
For SSS, you need three pairs of equal sides. Here we only know AD = BC and DB = DB (common side). We don't know if AB = CD, so we can't use SSS.
DB appears in both triangles △ADB and △CBD. Since it's the same line segment, DB = DB by the reflexive property - anything equals itself!
No, because we don't have two angles and the included side. We have one proven angle (from parallel lines), but we'd need another angle to be proven equal for ASA.
The middle part is always between the outer two parts!
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