For the quadrilateral ABCD, the equalities AB = BC = CD and AD = AC = BD are true.

For the quadrilateral ABCD, the equalities AB = BC = CD and AD = AC = BD are true. Find the smaller corner of the quadrilateral.

Note that triangles ACD and ABD are equal to each other on three sides, because AD – common side and AC = BD, AB = CD.

Hence it follows that the angles BAD = ABD = ADC = ACD = a.

Also note that triangles ABC and BCD are equal to each other on three sides, since AC = BD and AB = BC = CD.

This implies that BAC = BCA = CBD = CDB = b.

From the triangle ACD we have:

CAD + ACD + ADC = 180 °,

CAD = 180 ° – 2 * a.

Since the sum of the angles of a quadrilateral is 360 °:

BAD + ABD + CBD + BCA + ACD + ADC = 360 °,

a + a + b + b + a + a = 360 °,

2 * a + b = 180 °.

But from the triangle ABC we have:

BAC + ABC + BCA = 180 °,

b + 180 ° – a + b = 180 °,

a = 2 * b,

4 * b + b = 180 °, 5 * b = 180 °, b = 36 °.

a = 2 * b = 72 °.

Answer: 72 °.