Triangle Midsegment Theorem

Question

In △ABC, AC¯¯¯¯¯¯¯¯ and AB¯¯¯¯¯¯¯¯ are bisected at D and E, respectively. Determine the statement that could be used to show that DE¯¯¯¯¯¯¯¯∥CB¯¯¯¯¯¯¯¯.(1 point)

• If ∠AED≅∠EBC, then DE¯¯¯¯¯¯¯¯∥CB¯¯¯¯¯¯¯¯ because the alternate interior angles are congruent.

• If ∠ADE≅∠ACB, then DE¯¯¯¯¯¯¯¯∥CB¯¯¯¯¯¯¯¯ because the corresponding angles are congruent.

• If DE¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯, then △ABC≅AED by the side, side, side triangle congruence and DE¯¯¯¯¯¯¯¯∥CB¯¯¯¯¯¯¯¯ because they are the same line segment.

• If ∠DEB≅∠EBC, then DE¯¯¯¯¯¯¯¯∥CB¯¯¯¯¯¯¯¯ because the same-side interior angles are congruent

Answer 1

∵ D and E are mid-points AC and AB respectively.

∴ AD = 1/2 AC and AE = 1/2 AB.A

⇒ AD/AC = 1/2 and AE/AB = 1/2

⇒ AD/AC = AE/AB ...(1)

Now, in ΔADE and ΔACB,

∠DAE = ∠CAB (common angle)

And AD/AC = AE/BC (From (1))

∴ ΔADE ~ ΔACB (By SAS similarity Rule)

∴ ∠AED = ∠ABC (By C.P.C.T.)

∴ DE || CB (Because their corresponding are equal)

Option (B) is correct.