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In quadrilateral ABCD with diagonals BD and CA, $\angle ABD = 40^{ \circ } , \; \angle CBD = 70^{ \circ } , \; \angle ABD = 40^{ \circ } , \; \angle CDB = 50^{ \circ } , \; \angle ADB = 80^{ \circ } . \; Find \; \angle CAD .$ Read more…

Quadrilateral ABCD is inscribed in a circle. Let AB = a, BC = b, CD = c, DA = d, AC = p and BD = q. Prove Ptolemy’s second theorem that $p \cdot q = \frac{ad+bc}{ab+cd}$ .
It is easy question, just use $R = \frac{abc}{4S}$ a, b, c are sides of triangle and S is area of triangle.