3. Let $A B C$ be an isosceles triangle with base $B C=10$ and $A B=A C$. Construct isosceles triangles $D A B$ and $E A C$ externally on its two oblique sides, both similar to $A B C$, with $D A=D B$ and $E A=E C$. Given that $D E=45$, find the length of $A B$.
(A) 15
(B) 20
(C) $9 \sqrt{5}$
(D) 22.5
(E) It is not possible to determine it with the given data.
3. The answer is (A). Since the triangles $A B C, D A B, E A C$ are similar, it follows that the respective base angles are congruent; this means that $\widehat{D A B}=\widehat{A B C}=\widehat{A C B}=\widehat{C A E}$. Now we can calculate the measure of the angle $\widehat{D A E}=\widehat{D A B}+\widehat{B A C}+\widehat{C A E}=\widehat{A B C}+\widehat{B A C}+\widehat{A C B}=180^{\circ}$, which means $D, A, E$ are collinear. The triangles $D A B$ and $E A C$ are similar, and their bases $A B, A C$ have the same length, so they are also congruent and $D A=E A$; this, combined with the collinearity of $D, A, E$, leads to $D A=\frac{D E}{2}$. From the similarity of $D A B$ and $A B C$ we know that their sides are in proportion, so $\frac{D A}{A B}=\frac{A B}{B C}$ or $A B^{2}=D A \cdot B C$ and thus

$$
A B=\sqrt{D A \cdot B C}=\sqrt{\frac{D E \cdot B C}{2}}=\sqrt{\frac{45 \cdot 10}{2}}=15
$$