In quadrilateral abcd, the diagnosis intersect at point t. byron has used the alternate interior angles theorem to show that angle dac is congruent to angle bca and that angle bac is congruent to dac. which of the following can byron use prove that side ad is equal to side bc?

The rectangle ABCD and the congruent angles are shown in the diagram below.

We know that T is a midpoint.

That gives us the length of TD = length of TB, also TA = TC

Look for triangle ATD and BTC, we have so far that:
TD = TB and TA = TC
∠DAT = ∠BCT (properties of alternate angles)
∠ATD = ∠BTC (properties of opposite angles)

By the postulate SAS (Side Angle Side) we can deduce that AD = BC

TB = TD (Fourth option)