AB = BC, so triangle ABC then is isosceles.

Angle BAC = angle BCA = (1/2)(180 -112) = 34 degrees.

AE = ED, so triangle AED is isosceles too.

Angle ADE = angle BAC = 34 deg

Angle AED = 180 -34 -34 = 112 deg

Angle AEC is 180 deg.

Angle DEC = 180 -112 = 68 deg

This is the key to the problem: ED = DB. It must mean something.

Draw a line segment DF from point E to base AC such that DF is equal to ED.

Triangle EDF is isosceles.

Angle DFE = angle DEF = angle DEC = 68 deg

Angle EDF = 180 -68 -68 = 44 deg

Draw line segment BF.

DF = DB, so triangle BDF is isosceles.

Angle BDA is 180 deg. Angle DEF = 180 -44 - 34 = 102 deg

Angle DBF = angle DFB = (1/2)(180 -102) = 39 deg

Quadrilateral DBCA looks like a kite.

Any quadrilateral has 360 degrees for the sum of its interior angles.

(Interior angles = (n-2)(180) = (4-2)(180) = 360 deg.)

So, angle DFC = 360 -102 -112 -34 = 112 deg

That means angle DFC = angle DBC.

That means DBCA is a kite. That means DC is a diagonal, the longer diagonal and it bisects angle BCF.

Angle BCF = angle BCA = 34 deg

Therefore,

angle BCD = angle FCD = (1/2)(34) = 17 degrees. ----------answer.

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Another way, after finding angle DFC = angle DBC = 112 deg.

By triangle congruence SsA, meaning the given angle is opposite the longer side.

SSA is not 100% true for congruence. But there are special cases of SSA where congruence can be proven. One of them is SsA. Its commonly called "SsA Congruence Theorem"

(You can search the internet for that.)

CD = CD

DB = DF

angle DBC = angle DFC

SsA, so triangle CDB = triangle CDF = (1/2)(quadrilateral DBCA)

That means angle BCF is bisected by line segment CD.

Therefore, angle BCD = (1/2)(34) = 17 degrees ---------answer.