Question

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $\angle$DBC = 70${}^{\circ }$, $\angle$BAC = 30${}^{\circ }$, find $\angle$​BCD. Further, if AB= BC, find $\angle$​ECD.

Answer 1

$\angle$CDB = $\angle$BAC = 30\(^\circ) .....(i) (Angles in the same segment of a circle are equal)

$\angle$DBC = 70${}^{\circ }$ .....(ii)

In $\Delta$BCD,

$\angle$BCD + $\angle$DBC + $\angle$CDB = 180${}^{\circ }$ (Sum of all angles of a triangle is 180${}^{\circ }$)

$\angle$BCD + 70${}^{\circ }$ + 30${}^{\circ }$ = 180${}^{\circ }$ (using (i) and (ii))

$\angle$BCD = 180${}^{\circ }$ - 100${}^{\circ }$ = 80${}^{\circ }$ .....(iii)

In $\Delta$ABC,

Given: AB= BC

So, $\angle$BCA = $\angle$BAC = 30${}^{\circ }$.....(iv) (Angles opposite to equal sides of a triangle are equal)

Now, $\angle$BCD = 80${}^{\circ }$ from (iii)

$\angle$BCA + $\angle$ECD = 80${}^{\circ }$

30${}^{\circ }$ + $\angle$ECD = 80${}^{\circ }$

So, $\angle$ECD = 50${}^{\circ }$