If $AE$ and $BF$ are medians drawn to the legs of a right $\Delta ABC$, find the numerical value of $$\dfrac{(AE)^2 + (BF)^2}{(AB)^2}$$
What I Tried: First of all, I didn't quite understand what the question meant by numerical value. For example I took $AB = 5$ and $BC = 6$ in Geogebra, and the particular value I got to be was around $1.97$ an so on, but is this the answer for every right triangle? If yes, how? Also what does "numerical value" mean here?
Can anyone help me? Thank you.
On
Use the formulas for the medians and also the Pythagorean theorem.
https://www.math10.com/en/geometry/median.html
Say $a,b$ are the legs and $c$ is the hypotenuse.
$$m_a^2 = \frac{1}{4}\cdot(2c^2 + 2b^2-a^2)$$ $$m_b^2 = \frac{1}{4}\cdot(2c^2 + 2a^2-b^2)$$
Sum these two and you get:
$$m_a^2 + m_b^2 = \frac{1}{4}\cdot(4c^2 + a^2 + b^2) = \frac{1}{4}\cdot(5c^2) = \frac{5}{4}\cdot c^2 $$
The problem you are given asks you to find
$$\frac{m_a^2 + m_b^2}{c^2}$$
From the previous equality this is now obviously $$\frac{m_a^2 + m_b^2}{c^2} = \frac{5}{4}$$
In right $\triangle ACE$,
$$ AC^2 + (\frac{BC}{2})^2 = AE^2$$
In right $\triangle BCF$,
$$ (\frac{AC}{2})^2 + BC^2 = BF^2$$
Adding
$$ \dfrac{5}{4}(AC^2 + BC^2) = AE^2 + BF^2$$
$$ \therefore \dfrac{AE^2 + BF^2}{AB^2} = \dfrac{5}{4} = 1.25$$
As medians also form right triangles with legs, due to Pythagoras theorem it follows that this ratio is true for all right triangles.
According to question, $\angle C$ is right angle, $AB$ is hypotenuse. Probably this yielded the incorrect value of $1.97$.