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
.
- PROVE that the point of concurrency has coordinates
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 ratioif, 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.
- Find an internal point
- 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.