Finding Area of $\triangle ABF$ in a Right Triangle with Given Side Lengths and Median Intersection

Question

In right triangle $ABC$, we have $\angle ACB=90^{\circ}$, $AC=2$, and $BC=3$. Medians $AD$ and $BE$ are drawn to sides $BC$ and $AC$, respectively. $AD$ and $BE$ intersect at point $F$. Find the area of $\triangle ABF$.

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I aim to use the Shoelace Theorem to calculate the area of $\triangle ABF$, but I'm stuck on determining the coordinates of point $F$.

Could someone assist me in finding the coordinates of point $F$ so that I can proceed with calculating the area of $\triangle ABF$ using the Shoelace Theorem? Or is there another method altogether?

Answer 1

Great question. Please refer to my rough diagram $$\angle CAB = \angle CAD(=θ_1) + \angle BAD(=θ_2) = \tan ^{-1} \frac{CB}{AC} = \arctan (\frac{3}{2}) $$ $$θ_1=\arctan (\frac{CD}{AC})=\arctan(\frac{1.5}{2})$$ So $$θ_2=\angle CAB - θ_1 = 19.4400348282°$$ $$ \angle CBA = \angle CBE(=θ_3) + \angle EBA(=θ_4)= \arctan \frac{AC}{BC}=\arctan (\frac{2}{3}) $$ $$θ_3 = \arctan \frac{CE}{BC} = \arctan (\frac{1}{3})$$ $$θ_4=\angle CBA - θ_3= 15.2551187031°$$

In a triangle, if two angles(=$α_1,α_2$) and the length of side common to them(=$b$) are known, then Αrea of the triangle = $\frac{1}{2}\frac{b^2}{\cot α_1+\cot α_2}$

By Pythagorean theorem, $AB=\sqrt{13}cm$. So we know exactly the required amount of data to apply the formula given above. We know $\angle DAB,\angle EBA$ and $AB$.

Thus Area= $$\frac{AB^2}{2(\cot θ_2+\cot θ_4)}= \frac{13}{13} = 1 cm^2 $$

The area of the formula is an easy proof, and can be left as an exercise to the reader. Also you might not need a calculator by using the identity $$\arctan(x)-\arctan(y)=\arctan \bigg(\frac{x-y}{1+xy}\bigg) $$

Answer 2

Point $F$ is called the centroid of a circle. The medians of $\triangle ABC$ are always concurrent. Note: Centroids are usually denoted with the letter G.

In your image provided, draw a line containing $F$ perpendicular to $\overline {BC}$. Let's call the intersection point $G$. $\overline {BG}:\overline {GC}=\frac{2}{3}:\frac{1}{3}$.

Assuming $C$ = ($0, 0$), the $x$-coordinate of $F=1$ and the $y$-coordinate of $F=\frac{2}{3}$. $F$ is located at ($1, \frac{2}{3}$).

Answer 3

Also I would like to mention another very easy process, using data given by @Wasu Chanyasubkit. You know the coordinates of $A(0,2),B(3,0),C(1,\frac{2}{3})$. You can apply the formula for area of a triangle between 3 coordinates.

$$A=\frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))$$

To get $A=1 cm^2$

Answer 4

The 3 medians divide $\Delta ABC$ into $6$ triangles of equal area.

Since $\Delta ABF$ consists of $2$ small triangles, area of $\Delta ABF$ equals to one third of the area of $\Delta ABC$.

Since area of $\Delta ABC$ equals $3$, area of $\Delta ABF$ equals $1$.