# Mix and match cevians

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

In a triangle, a cevian is a line segment from a vertex to the opposite side. A median is a special cevian that divides the opposite side in the ratio . Our aim in this post is to modify the ratios in which cevians divide the opposite sides, in such a way that concurrency is forced.

## 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.

We can also find points and explicitly, and use the fact that cevians and are divided in the ratio at 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.

1. 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 .
2. has vertices at . Assuming cevians divide sides in the ratios , respectively, and they concur at the point , determine and .
3. has vertices at . Determine the lengths of cevians which divide sides in the ratios , respectively.
4. 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.)
5. 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.)
6. (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.
7. 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 ).
8. 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.
9. 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.
10. 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.