Question

an isosceles triangle, \\( \triangle a b c \\) with \\( \overline{a b} \cong \overline{c b} \\) is shown. points \\( d, e \\), and \\( h \\) are the midpoints of \\( \overline{a b}, \overline{c b} \\), and \\( \overline{a c} \\), respectively. points \\( f \\) and \\( g \\) are the midpoints of \\( \overline{d h} \\) and \\( \overline{e h} \\), respectively. what is the ratio of the area of \\( \triangle a b c \\) to the area of \\( \triangle f g h \\)?
\\( \bigcirc \\) a. \\( 4: 1 \\)
\\( \bigcirc \\) b. \\( 8: 1 \\)
\\( \bigcirc \\) c. \\( 16: 1 \\)
\\( \bigcirc \\) d. \\( 64: 1 \\)

Explanation:

Step1: Use the mid - point theorem

By the mid - point theorem, \(DE\parallel AC\), \(DE=\frac{1}{2}AC\), \(DH\parallel BC\), \(DH = \frac{1}{2}BC\), \(EH\parallel AB\), \(EH=\frac{1}{2}AB\).
Since \(\triangle ABC\) is isosceles (\(AB = CB\)), and \(D\), \(E\), \(H\) are mid - points.
The ratio of the sides of \(\triangle DEH\) to \(\triangle ABC\) is \(1:2\) (because of mid - point theorem).

Step2: Consider the similarity of triangles

If two similar triangles have a side - ratio \(k\), the ratio of their areas is \(k^{2}\).
For \(\triangle DEH\) and \(\triangle ABC\), \(k=\frac{1}{2}\), so \(A_{\triangle DEH}=\frac{1}{4}A_{\triangle ABC}\).
Now, \(F\) and \(G\) are mid - points of \(DH\) and \(EH\).
For \(\triangle FGH\) and \(\triangle DEH\), the side - ratio \(k'=\frac{1}{2}\) (by mid - point theorem for \(\triangle DEH\)).
The ratio of the areas of \(\triangle FGH\) and \(\triangle DEH\) is \((\frac{1}{2})^{2}=\frac{1}{4}\), so \(A_{\triangle FGH}=\frac{1}{4}A_{\triangle DEH}\).

Step3: Find the ratio of \(A_{\triangle ABC}\) and \(A_{\triangle FGH}\)

Since \(A_{\triangle DEH}=\frac{1}{4}A_{\triangle ABC}\), then \(A_{\triangle FGH}=\frac{1}{4}\times\frac{1}{4}A_{\triangle ABC}=\frac{1}{16}A_{\triangle ABC}\).

Answer:

C. \(16:1\)