We (can always) construct two cevians that concur with a median.

## Ratios 1:2, 1:1, 1:2

In the diagram below, let cevian divide side such that ; let be a median, so that ; let cevian divide side in the ratio .

We show that at the point of concurrency, median is bisected in the ratio , while cevians and are divided in ratios . (Compare this with the manner in which the centroid divides the three medians of a triangle.)

Using the given ratios, PROVE that are concurrent.

Just like proving that the three medians of a triangle are concurrent, this question belongs at the simplest level of simplicity. We have:

By the converse of Ceva’s theorem, we conclude that are concurrent.

In the diagram below, the small letters represent the areas of the triangles in which they are located. PROVE that , given that , , .

Let’s do this in four steps:

- First, we have . Consider and . They have the same altitudes, measured from their common vertex . Since their base lengths and are also equal, their areas must be equal as well. This gives .
- Next we have and . Consider and . They have the same altitudes, measured from their common vertex . Since , the area of is twice that of so . For the same reason we have , when and are compared.

Our diagram simplifies to: - We now show that . Consider and . They have the same altitudes, measured from their common vertex . Since , these two triangles have the same areas, so . This gives .
- Lastly, we show that . Consider and . They have the same altitudes, measured from their common vertex . Since , the area of is twice the area of . Thus, , and so . Since from the previous step, we get as desired.

Suppose that , , and in the diagram below:

PROVE that at the point of concurrency, is bisected in the ratio , is divided in the ratio and is divided in the ratio .

Let the point of concurrency be and let the area of be denoted by . Then, in view of the previous example, the simplified area diagram, in terms of , is:

Consider and . They have the same areas, namely . Since they also have the same altitudes (measured from their common vertex ), it must be the case that their base lengths are also equal. Thus .

Now consider and . They have the same altitudes, measured from their common vertex . The area of is , while the area of is . Thus their base lengths and must be in the ratio of their areas, so .

Similarly we have .

has vertices at , , and . Cevians divide sides in the ratios , respectively. Determine the coordinates of the point of concurrency.

Note that cevian is a median. So it is bisected at the point of concurrency, in view of the given ratios. This means that the midpoint of is the point of concurrency. Since is the midpoint of , it is the point . Then we take the midpoint of and to get as the point of concurrency.

- Since divides in the ratio , it is the point . Then the point of concurrency divides in the ratio , measured from vertex . So it is the point , as before.
- Since divides in the ratio , it is the point . The point of concurrency divides cevian in the ratio , measured from . So it is the point , as before.

So we obtain the same point , but the last two approaches involved more steps.

## Ratios m:n, 1:1, m:n

This is a slight generalization of the previous examples.

In the diagram below, the small letters represent the areas of the triangles in which they are located. PROVE that , given that , , .

As before, we do this in steps:

- First we have , because and both and have the same height.
- Next we have . This follows because and both and have the same altitudes, so their areas are proportional to their base lengths. Similarly, we have . At this stage, our simplified diagram is:
- Next we show that . Consider and . Since , these two triangles have the same areas, so . That is, . Re-arranging this gives , or . Since both and are positive (we’re dealing with internal division), we must have , so . Let’s adjust the diagram a bit:
- Lastly we show that . Consider and . They have the same altitudes, measured from their common vertex . Since , their areas follow the same ratio, meaning that .
Thus we get as desired.

Suppose that , , and in the diagram below:

PROVE that at the point of concurrency, is divided in the ratio , is divided in the ratio , and is divided in the ratio .

Let the point of concurrency be and let the area of be square units. Then, in terms of , we have the following area diagram:

Consider and . They have the same altitudes, measured from their common vertex . So the ratio of their areas gives the ratio of their base lengths. In other words, .

To show that , consider and . They have the same altitude, measured from their common vertex . So the ratio of their areas gives the ratio of their base lengths.

Similarly, .

PROVE that the centroid of a triangle divides the three medians in the ratio .

In the case of three medians, we have . So, using this in the previous example:

- the median is divided in the ratio
- the median is divided in the ratio
- the median is divided in the ratio

In , is a median. In what ratio should cevians and divide sides and so that the point of concurrency and the centroid trisect median ?

Let be the centroid, and let the point of concurrency be . Then divides median in ratio , measured from vertex ; that is, . Since is to be trisected, we want , or . From the previous example, the median is divided in the ratio at the point of concurrency. Therefore , or . So cevian should divide side in the ratio and cevian should divide side in the ratio .

For any right triangle with legs parallel to the coordinate axes, PROVE that the slopes of its three medians form a geometric progression with common ratio .

For simplicity, fix the vertices at , , and ; a little modification will be needed for the general case.

We now calculate the slopes of medians :

These form a geometric progression with common ratio of .

(There’s something interesting here. In our first post of this year, we saw that if the slopes of the sides of a triangle form a geometric sequence with as common ratio, then the triangle contains a vertical median and a horizontal median. Now, the above example says that if a triangle contains a vertical side and a horizontal side, then the slopes of its three medians form a geometric progression with a common ratio of . Did you notice something? See exercise 5 at the end.)

Give an example of a triangle in which one side, one median, and another cevian have slopes forming an arithmetic progression.

Consider below, in which is a median, is a cevian:

The slopes of are , respectively. They form an arithmetic progression with a common difference of .

(Notice that if a third cevian is drawn from vertex and is concurrent with cevian and median , then it is an altitude. This is a good example of “mixing” and “matching” cevians as used in our post’s title. Mixing means that they are different; matching means that they are concurrent.)

## Takeaway

The ratios we considered in this post () were for convenience. There are other ratios that can still ensure concurrency.

## Tasks

- Let be the vertices of . If there are cevians which divide sides in the ratios , respectively:
- PROVE that the point of concurrency has coordinates ;
- Deduce that the centroid of is the point .

- has vertices at . Assuming cevians divide sides in the ratios , respectively, and they concur at the point , determine and .
- has vertices at . Determine the lengths of cevians which divide sides in the ratios , respectively.
- Let be a right triangle in the first quadrant, with vertices , , and . For , PROVE that there are three cevians whose slopes form a geometric progression with common ratio .

(Compare this with Example 9.) - Let be the vertices of . Let the slopes of medians through be denoted by , respectively. Suppose that (in that order) form a geometric progression with a common ratio of . PROVE that:
- ;
- .

(In view of Example 9, conclude that the slopes of the three medians of a triangle form a geometric progression with common ratio if, and only if, the triangle is a right triangle with legs parallel to the coordinate axes.)

- (Opposite slopes) The right triangle with vertices at , , has very nice properties:
- Find an internal point on such that the slope of cevian is the negative of the slope of cevian , and PROVE that these two cevians intersect at ;
- Find an internal point on such that the slope of cevian is the negative of the slope of cevian , and PROVE that these two cevians intersect at ;
- If a cevian from and a cevian from have opposite slopes, is their point of intersection always of the form for some ?
- For any point between and , PROVE that there is a point between and such that the slopes of cevians and are negatives of each other;
- For any point between and , PROVE that there is a point between and such that the slopes of cevians and are negatives of each other.

- Consider the right triangle with vertices at , , :
An arbitrary point on can be given by the coordinates , where . PROVE that:

- if and only if ;
- if and only if ;
- for and , cevians and have opposite slopes;
- for and , cevians and have opposite slopes;
- a cevian from and a cevian from intersect at if and only if they have opposite slopes (and );
- a cevian from and a cevian from intersect at if and only if they have opposite slopes (and ).

- Let be the vertices of . Suppose that the slopes of sides and are negatives of each other, and that the slopes of medians through and are also negatives of each other. PROVE that:
- (so form an arithmetic sequence)
- (so is isosceles)
- the median through vertex is vertical.

- PROVE that if the slopes of the equal sides of an isosceles triangle are negatives of each other, then the -coordinates form an arithmetic progression.
- For an integer , let be the vertices of . PROVE that a cevian from vertex divides side in the ratio if and only if it is an altitude.