Prove if angle bisectors of a pair of opposite angles of quad. meet on diagonal made by remaining points then the remaining points will do same

Question

In a quadrilateral $ABCD$, the bisectors of $\angle A$ and $\angle C$ meet on $BD$, prove that the bisectors of $\angle B$ and $\angle D$ meet on $AC$.

Just give a hint in comments first please. If I fail to get the answer then I will tell you and then give the answer.

Answer 1

In quadrilateral $ABCD$, bisector of $\angle A$ meets $BD$ at $F$, bisector of $\angle B$ meets $AC$ at $E$, bisector of $\angle C$ meets $BD$ at $I$, bisector of $\angle D$ meets $AC$ at $G$.

By angle bisector theorem, $$\frac{AB}{AD}=\frac{BF}{FD}$$ $$\frac{BC}{CD}=\frac{BI}{ID}$$ Now, for the condition given to be true $F=I$ Therefore, $$\frac{AB}{AD}=\frac{BC}{CD}$$ By rearranging it we get, $$\frac{AB}{BC}=\frac{AD}{CD}$$ which again by angle bisector theorem means, $$\frac{AE}{EC}=\frac{AG}{GC}$$ This implies $G=E$ thus proving our statement.