Answer:
[tex]\dfrac{1}{3}[/tex]
Step-by-step explanation:
If points M and N are the midpoints of sides AC and BC of △ABC, then segment MN is a midline segment of the triangle ABC. By the triangle midline theorem, MN║AB and is half of AB.
Since
- angle C is common;
- MN║AB, then angles A and M are congruent and angles B and N are congruent,
then triangles ABC and MNC are similar by AAA theorem with [tex]\dfrac{1}{2}[/tex] factor of similarity. Thus,
[tex]A_{\triangle MNC}=\dfrac{1}{4}A_{\triangle ABC}.[/tex]
The area of the quadrilateral ABNM is
[tex]A_{ABNM}=A_{\triangle ABC}-A_{\triangle MNC}=A_{\triangle ABC}-\dfrac{1}{4}A_{\triangle ABC}=\dfrac{3}{4}A_{\triangle ABC}.[/tex]
Then
[tex]\dfrac{A_{\triangle MNC}}{A_{ABNM}}=\dfrac{\frac{1}{4}A_{\triangle ABC}}{\frac{3}{4}A_{\triangle ABC}}=\dfrac{1}{3}.[/tex]
