gobathegreat wrote: Let $ABC$ be an isosceles triangle such that $\angle BAC = 100^{\circ}$. Let $D$ be an intersection point of angle bisector of $\angle ABC$ and side $AC$, prove that $AD+DB=BC$ Let $E$ be a point on $BC$ such that $BD=BE$ $\angle BDE=\angle BED=80$ $\angle CDE=180-60-80=40=\angle DCE$ $DE=EC$ $\angle BAD+\angle BED=180$ $ABED$ concyclic $\angle DEA=\angle DBA=20=\angle DBE=\angle DAE$ $AD=DE$ $AD+DB=DE+DB=EC+DB=EC+BE=BC$

angle BAC=100° AB=AC angle ABC=angle ACB=40° angle ABD=angle DBC=20° In ∆ABD, sin 100°/ BD=sin 20°/AD= sin 100°+sin 20°/ BD+AD BD+AD / BD=sin 100°+sin 20° / sin 100° =2sin 60° cos 40°/ sin 100° =2sin 60° cos 40° sin 40°/ sin 100° sin 40° =sin 80° sin 60° / sin 80° sin 40° =sin 60°/ sin 40°...(1) In ∆ BDC, angle BDC=120° BC/ BD=sin 120°/ sin 40°=sin 60°/ sin 40°...(2) BD+AD / BD=BC/ BD So, BC=BD+AD @KRISHIJIVI