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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)?

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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.

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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.
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  • $\begingroup$ Thanks, good answer !! $\endgroup$ – Mattiu May 16 at 4:02

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