Question

In a cyclic quadrilateral ABCD, if $(\angle B-\angle D)={60}^{\circ }$, show that the smaller of the two is ${60}^{\circ }$.

Answer 1

ABCD is a cyclic qudrilateral

∠B-∠D = 60${}^{\circ }$ ------ [1] (given)

∠B+∠D = 180${}^{\circ }$ ------ [2] (opp ang of a quad is 180${}^{\circ }$ )

from equation 1 and 2

∠B-∠D+∠B-∠D = 180${}^{\circ }$+60${}^{\circ }$

∠B+∠B = 240${}^{\circ }$

2∠B = 240${}^{\circ }$

∠B = 240${}^{\circ }$/2

∠B = 120${}^{\circ }$

∠B+∠D = 180${}^{\circ }$

120${}^{\circ }$+∠D = 180${}^{\circ }$

∠D = 180${}^{\circ }$-120${}^{\circ }$

∠D = 60${}^{\circ }$

thus, smaller angle of two that is ∠D = 60${}^{\circ }$