Area of triangle given two medians and one side.

Question

In triangle $ABC$,point $D$ is the midpoint of $BC$and point $E$ is the midpoint of $AC$.If $AD=6$,$BE=9$,and $AC=10$ then find the area of triangle $ABC$.

I've solved this problem in two ways:

$1)$ by constructing the third median and calculating the area of one of the triangle determined by the intersection of the medians then multipy by six.

$2)$ by calculating the lengths of the other two unknown sides and then applying Heron's Formula.

Now what i am asking is if someone can come up with another solution based on geometric methods.

P.S:you may ask the why of this question and the answer would be that i like to see many creative solutions to one problem,and see what is the most beautiful one.

Thanks in advance

Answer 1

Let $BE$ amd $AD$ meet at G, the centroid. It is a well-known property of the centroid that it divides a median in the ratio $2:1$. Therefore $AG=4$ and $GE=3$. Therefore $AEG$ is a $3-4-5$ i.e. right-angled triangle whose area is $6$. Therefore the total area is $36$

Answer 2

I think @David Quinn's answer is excellent and to the point, but it may be difficult for our novice readers to understand. Therefore, I'd like to give a detailed answer, which can be easy to understand by highschoolers. Following image shows the detailed data given by the question:

As shown in the image, $AC = 10$, $AD = 6$, and $BE = 9$. Since $D$ and $E$ are midpoints of $BC$ and $AC$, $AE = CE = 5$. Suppose $AD$ and $BE$ crossed at point $G$. Then, if you extend $CG$ until it meets $AB$ at point $F$, then point $F$ would be the midpoint of $AB$ (a property of triangle and G is called the center of gravity of the triangle). Thus, $$\frac{AG}{GD} = \frac{BG}{GE} = \frac{CG}{GF} = \frac{2}{1}$$

Since $AD = 6$ and $BE = 9$, $AG = 4$, $GD = 2$, $BG = 6$, and $GE = 3$. Thus, $\triangle AGE$ is a pythagorean triangle $(AG^2 + GE^2 = 4^2 + 3^2 = 5^2 = AE^2)$, hence $AD$ and $BE$ are perpendicular to each other.

If the area of $\triangle AGE$ is $\alpha$, then $\alpha = \frac12 \cdot 4 \cdot3 = 6$. Let's consider the 6 triangles around point $G$. If the area of a triangle is $A_\triangle$, then:

$$A_{\triangle AGE} = A_{\triangle CGE} = \alpha = 6 \qquad A_{\triangle BGD} = A_{\triangle CGD} = \beta \qquad A_{\triangle AGF} = A_{\triangle BGF} = \gamma $$

Since $A_{\triangle ABE} = A_{\triangle CBE}$, $2 \beta + \alpha = 2 \gamma + \alpha$, and hence, $ \beta = \gamma $. Similarly, considering $A_{\triangle ABD} = A_{\triangle ACD}$, you can show that $ \alpha = \gamma $. Therefore, $ \alpha = \beta = \gamma = 6$. Accordingly,

$$A_{\triangle ABC} = 2\alpha + 2\beta + 2\gamma = 6 \times \alpha = 36$$