Question

practice 3
in this diagram, lines ac and de are parallel, and line dc is perpendicular to each of them.
\\( \

$$\begin{array} { l } { \\overleftrightarrow { a c } \\parallel \\overleftrightarrow { d e } } \\\\ { \\overleftrightarrow { d c } \\perp \\overleftrightarrow { d e } } \\\\ { \\overleftrightarrow { d c } \\perp \\overleftrightarrow { a c } } \\end{array}$$

\\)
what is a reasonable estimate for the length of side be?
a \\( \frac { 1 } { 3 } \\)
b 1
c \\( \frac { 5 } { 3 } \\)
d 5

Explanation:

Step1: Use the property of similar triangles

Since \(AC\parallel DE\), \(\triangle ABC\sim\triangle DBE\).
Let \(BE = x\). Then \(DB=DC - BC\). Given \(DC\perp AC\), \(DC\perp DE\), \(AC\parallel DE\), \(AB = 5\), \(BC=4\). Let \(DC = h\), \(AC = k\).
By the similarity of \(\triangle ABC\) and \(\triangle DBE\), we have \(\frac{AB}{BE}=\frac{BC + BD}{BD}\) (or use the ratio of corresponding sides).
Another way: Let's assume the ratio of similarity. The length of \(AB = 5\), and \(BC = 4\). Let \(BD=h - 4\) (where \(h = DC\)).
Since \(\triangle ABC\sim\triangle DBE\), we know that \(\frac{AB}{BE}=\frac{BC}{BD}\) (by the property of similar - triangles \(\frac{\text{hypotenuse}_1}{\text{hypotenuse}_2}=\frac{\text{leg}_1}{\text{leg}_2}\)).
Let \(BE=x\), then \(\frac{5}{x}=\frac{4}{BD}\). Also, from the Pythagorean theorem in \(\triangle ABC\), if we assume \(AC\) and \(DC\) are the legs of right - angled triangles.
Since \(AC\parallel DE\), we can use the fact that the triangles are similar. The ratio of the sides of \(\triangle ABC\) and \(\triangle DBE\):
Let's consider the fact that if we assume the length of \(AC\) and \(DE\) (the non - perpendicular sides) are in proportion.
We know that \(AB = 5\), \(BC = 4\), assume \(BD = 1\) (because \(DC\) is a common perpendicular, and if we consider the proportion).
By the similarity of right - angled triangles \(\triangle ABC\) and \(\triangle DBE\) (\(\angle C=\angle BDE = 90^{\circ}\), \(\angle ABC=\angle DBE\) (common angle)), \(\frac{AB}{BE}=\frac{BC}{BD}\).
If \(AB = 5\), \(BC = 4\), and assume \(BD = 1\) (because \(DC\) is a line segment and \(BC = 4\), and \(AB\) is the hypotenuse of \(\triangle ABC\)).
Substitute into \(\frac{AB}{BE}=\frac{BC}{BD}\), we get \(\frac{5}{BE}=\frac{4}{1}\), \(BE=\frac{5}{4}\approx1\).

Answer:

B. 1