Acute Angle Measure

$2[AEF]=[ABC]$ . By using Sine Formula for the area of a triangle, the equation becomes $2\cdot \frac{AE\cdot AF \sin \angle BAC}{2} =\frac{AC\cdot AB \sin \angle BAC}{2}$ , which simplifies to $2AE\cdot AF=2AC\cdot AB$ . We have $AF=AC\cos \angle BAC$ and $AE=AB\cos \angle BAC$ . Substituting this in the equation above, $2AC\cdot AB\cos^2 \angle BAC=AC\cdot AB$ . Hence, $2\cos^2 \angle BAC=1$ , which gives $\angle BAC=45^\circ$

Let the angle to be calculated be x $$2[AEF]=[ABC]$$ $$2(0.5(AE)(AF)sin(x))=(0.5)(AB)(AC)sin(x)$$ $$2(AE)(AF)=(AB)(AC)$$ $$\angle AED = 90^\circ \Rightarrow AE = ABcos(x)$$ $$\angle AFC = 90^\circ \Rightarrow AF = ACcos(x)$$ Substituting these two into the equation above, we get $$2(AB)(AC)(cos^2 (x)) = (AB)(AC)$$ $$cos^2 (x) = 1/2$$ $$cos (x) = 1/ \sqrt(2)$$ $$x < 90^\circ \Rightarrow x = cos^{-1} (1/ \sqrt(2)) = 45^\circ$$

Triangle ABC is an acute triangle. E be the foot of the perpendicular from B to AC, F be the foot of the perpendicular from C to AB. Join EF Since triangle ABC & AEF are similar as, \angle AEF = \angle ABC , \angle AEF = \angle ACB Therefore, \frac {[AEF]}{[ABC]} = (\frac {AE}{AB})^2 = (\frac {AF}{AC})^2 = (\frac {FE}{BC})^2

According to question, \frac {[AEF]}{[ABC]} = \frac {1}{2} (\frac {AE}{AB})^2 = \frac {1}{2} \frac {AE}{AB} = \frac {1}{\sqrt{2}}

Let \angle BAC be \theta , then, \sin \theta = \frac {AE}{AB} \sin \theta = \frac {1}{\sqrt{2}} \sin 45^\circ = \frac {1}{\sqrt{2}} \angle BAC = 45 ^ \circ

Let $G$ be the foot of the perpendicular from $F$ to $AC$ . Then $2[AEF] = [ABC]$ is equivalent to $|AE||FG| = \frac{1}{2} |BE||AC|$ , or $\frac{|FG|}{|AC|} = \frac{|BE|}{2|AE|}$ .

Let $\theta$ be the measure of angle $BAC$ . (This is the desired angle.) Then using basic trig. definitions (and the figure), we can write: $\begin{aligned} \tan(\theta) &= \frac{|BE|}{|AE|} \\ \cos(\theta) &= \frac{|AF|}{|AC|} \\ \sin(\theta) &= \frac{|FG|}{|AF|} \end{aligned}$

Combining the last two gives: $\frac{|FG|}{|AC|} = \cos(\theta)\sin(\theta)$ .

Then our equation area above can be written as $\begin{aligned} \cos(\theta)\sin(\theta) &= \frac{1}{2} \tan(\theta) \\ \cos^2(\theta) &= \frac{1}{2} \\ \cos(\theta) &= \frac{1}{\sqrt{2}}. \end{aligned}$

So $\theta = 45$ (in degrees).

First we note that $\triangle AEF \sim \triangle ABC$ . So we have $\dfrac{AF}{AB} = \dfrac{AE}{AC} = \dfrac{FE}{BC}$ Now, $\dfrac{[AEF]}{[ABC]} = \dfrac{AE \times AF}{AB \times AC} = (\dfrac{EF}{BC})^2$ By our given condition, we have,

$\dfrac{EF}{AB} = \dfrac{1}{\sqrt{2}}$ But EF is a side of the orthic triangle, and so we have $EF = a \cos \angle A$

Thus, $\dfrac{ a \cos \angle A}{a} = \dfrac{1}{\sqrt{2}}$

Therefore, $\cos\angle A = \dfrac{1}{\sqrt{2}}$ . Thereofre, $\angle A = 45^{\circ}$ as it is an acute angled triangle.

1/2=[AEF]/[ABC]=(AF AE)/(AB AC) So we can know AC/AF=AB/AE=sqrt(2) The measure is 45 degrees