Question

Points M and N are the midpoints of sides AC and BC of △ABC. Prove that the ratio of the area of △MNC to the area of ABNM is 1:3.

Answer 1

`(1)/(3)`

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
`(1)/(2)` factor of similarity. Thus,

`A_(\triangle MNC)=(1)/(4)A_(\triangle ABC).`

The area of the quadrilateral ABNM is

`A_(ABNM)=A_(\triangle ABC)-A_(\triangle MNC)=A_(\triangle ABC)-(1)/(4)A_(\triangle ABC)=(3)/(4)A_(\triangle ABC).`

Then

`(A_(\triangle MNC))/(A_(ABNM))=((1)/(4)A_(\triangle ABC))/((3)/(4)A_(\triangle ABC))=(1)/(3).`