Nice ratios

Here is a very simple but nice result.
Let ABC be a triangle and let D the midpoint of BC. Let E be the midpoint of AD and let F be the point where CE meets AB. Then AF : FB = 1 : 2 and FE : EC = 1 : 3.
nice ratios 1
This is appealing (to me at least) because this simple configuration produces the ratios 1 : 1, 1 : 2 and 1 : 3, and these properties can be proved in many ways; I will illustrate three such ways here; the first two involve simple constructions. The idea is to add points and lines in order to manufacture similar triangles – then parallel lines and enlargement arguments can be used. The third considers areas.

Method 1
Let M be the midpoint of BD.
nice ratios 2.png
E is the midpoint of AD so triangle MED enlarges to triangle BAD, centre D and scale factor 2 – so ME is parallel to AB and ME = 1/2 × AB.
Since ME is parallel to FB, triangle CME enlarges to triangle CBF, centre C. Since MD is half of BD and BD is half of BC, MC = 3/4 × BC so the scale factor is BC/MC = 4/3; thus CF = 4/3 × CE so FE : EC = 1 : 3.
Also, by the same enlargement, FB = 4/3 × ME = 4/3 × 1/2 × AB = 2/3 × AB, so AF : FB = 1 : 2.

Method 2
This method is personally my favourite.
Let P be the point on AD extended such that D is the midpoint of EP.
nice ratios 3.png
Since D bisects BC and EP, BECP is a parallelogram.
So EC is parallel and equal to BP. So FE is parallel to BP. Hence there is an enlargement carrying triangle AFE to triangle ABP, centre A. This enlargement has scale factor 3, since AP = 3 × AE.
Hence AB = 3 × AF so AF : FB = 1 : 2.
Also BP = 3 × FE by the same enlargement; but BP = EC so FE : EC = 1 : 3.

Method 3
We are going to use the fact that area of triangle = base × perpendicular height. If two triangles have the same perpendicular height, then their areas are in the ratio of their bases.
AQA4
This works both ways: if we know the ratio of the areas, we know the ratio of the bases, and vice versa. So, we can apply this to our diagram:
nice ratios 4
Let [XYZ] denote the area of a triangle with vertices X, Y and Z. Let [ABC] = 12T.
BD = DC so [ABD] = [ADC]; hence [ABD] = [ADC] = 6T.
Also AE = ED so [ABE] = [BED] = 3T and [AEC] = [CED] = 3T.
In other words, AD, BE and CE divide the triangle into quarters.
Let AF : FB = 1 : k. Then [AFC] : [BFC] = 1 : k.
However, [AFE] : [BFE] = 1 : k also.
Hence, subtracting gives [AEC] : [BEC] = 1 : k.
However [BEC] = [BED] + [CED] = 6T so [AEC] : [BEC] = 3T : 6T = 1 : 2
Hence k = 2 so AF : FB = 1 : 2.
Therefore, [AFE] = 1/3 × [ABE] = T.
So FE : EC = [AFE] : [AEC] = T : 3T = 1 : 3.

There are other methods – let me know if you find any particularly elegant ones!

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s