BTW when you see a right triangle a thing I always do is imagine it is inscribed in a circle with the hypotenuse as the diameter. That's just what I do, no need to do what I do. But is always nice to do what I do, because doing what I do makes me feel like people exist. I like when people exist, especially people who do what I do.
Anyways, draw a segment from \(A\) to to a point on \(BC\) in such it way that it is perpendicular to the hypotenuse. We call the point of intersection on the hypotenuse \(L\).
This means \(AL\perp{BC}\).
- BC = 5 because it is a 3-4-5 triangle.
By AA similarity, we know that triangle \(ALC\) is similar to triangle \(ABC\). To find AL, we write the following proportion:
\(\frac{AL}{3}=\frac{4}{5}\)
Solving for AL:
1. \(5AL=12\)
2. \(AL=\frac{12}{5}\)
Now here is how you solve the rest:
1. Consider point \(M\), that is the midpoint of \(BC\). Because it is the midpoint, we know that \(CM=\frac{5}{2}\)
2. Use the pythagorean theorem to find \(CL\) hint:\(\sqrt{AC^2-AL^2}\)
3. Now that you found \(CL\), we can find \(LM\) by evaluating \(CM-CL\)
4. Since you know \(AL\) and \(LM\), you can easily find \(AM\) by pythagorean theorem. hint: \(\sqrt{AL^2+LM^2}\)
Yay
