In triangle $ABC,$ $M$ is the midpoint of $\overline{AB}.$ Let $D$ be the point on $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC,$ and let the perpendicular bisector of $\overline{AB}$ intersect $\overline{AD}$ at $E.$ If $AB = 44,$ $AC = 30,$ and $ME = 10,$ then find the area of triangle $ACE.$
Hi Guest!
First, let's draw the triangle and all of the given:
Now, to find the area of triangle AEC, we already have the base (AC=30), we need the height, that is, the distance from E to AC.
Notice: AM=22 (44/2, as M is the midpoint of AB). and ME is given to be 10.
So with trigonometry: \(tan(\theta)=\frac{10}{22}=\frac{5}{11}\)
And, by pythagoras: \(AE=\sqrt{22^2+10^2}=\sqrt{584}\)
If \(tan(\theta)=\frac{5}{11} \implies sin(\theta)=\frac{5}{\sqrt{146}}\) (Or simply using the triangle above, sin(theta) = opposite/hypotenuse which is ME/AE).
Next, in the triangle with the "Other theta" That is, the triangle having "h"
\(sin(\theta)=\frac{h}{\sqrt{584}} \iff \frac{5}{\sqrt{146}}=\frac{h}{\sqrt{584}}\)
Therefore, \(h=10\)
So, the area of triangle ACE = \(\frac{30*10}{2}=150\)
