# What statement is correct in the problem about similarity

I have this statement:

In the rectangle ABCD of Figure 24, AE is bisector of $$\angle BAD$$ and DB is diagonal. If AD = a and AB = b, which (is) of the following statements is (are) true (s)?

My current development is:

i) According to interior bisector theorem: $$\frac{FB}{FD} = \frac{b}{a}$$, also $$\angle BAE \cong ,\angle FED$$ and $$\angle AFB \cong \angle EFD$$, so $$\triangle AFB \sim \triangle DEF$$, and the reason of similarity $$\frac{ \triangle AFB}{ \triangle DEF} = \frac{b}{a}$$ and the area will be the square of that reason, and as they ask me the inverse reason, it will be: $$\frac{a^2}{b^2}$$, thus this statement is correct. Also i can note that $$DE = b*a/b = a$$

The problem is that I have not managed to do anything useful with the 2 following statements, I am learning this recently on my own and I am very stagnant, thanks in advance.

Note that $$AE$$ is the angle bisector of $$\angle DAB$$, so the angles $$\angle DAE$$, $$\angle EAB$$, $$\angle AED$$ have each $$45^\circ$$.

We use now the following simple facts.

• If two triangles are similar, with length similarity factor $$k$$, then the areas are in the proportion $$k^2$$. In our case $$\Delta FED$$ and $$\Delta FAB$$ are similar.
• If two triangles have the same height (corresponding to appropriate bases), then the proportion of the areas is the proportion of the corresponding bases. In our case $$\Delta AFD$$ and $$\Delta DFE$$ have the same height from $$D$$ corresponding to the line of the bases $$AF$$, and $$FE$$, same line. Also, $$\Delta AFD$$ and $$\Delta ABD$$ have the same height from $$A$$ corresponding to the line of the bases $$FD$$, and $$BD$$, same line.

So the three proportions of areas are:

• first, $$(DE:AB)^2=(AD:AB)^2=(a:b)^2=a^2:b^2$$,
• second, $$AF:FE=b:a$$,
• third, $$FD:BD=FD:(FD+FB)=DE:(DE+AB)=a:(a+b)$$, derived proportions were used.
• Thanks, good answer !! – Mattiu May 16 at 4:02