ABCD – isosceles trapezoid (AD-base). Perimeter ABCD = 27, AD-BC = 5, BCA angle equals DCA angle. Find BC and AD.

Since, by condition, the angle BCA = DCA, then AC is the bisector of the angle BCD.

Angle CAD = BCA as criss-crossing angles at the intersection of parallel straight lines BC and AD secant AC, then angle DSA = CAD, which means triangle ACD is isosceles C = AD. Then AB = AD.

Let the length of the base BC = X cm, then, by condition, AD = (X + 5) cm.

Then the perimeter of the trapezoid is: P = X + (X + 5) + (X + 5) + (X + 5) = 27.

4 * X = 27 – 15 = 12 cm.

X = BC = 12/4 = 3 cm, then AD = 3 + 5 = 8 cm.

Answer: BC length is 3 cm, AD length is 8 cm.