Today I've encountered a question like the following;
$ABC$ is an acute angled triangle $m(\widehat{A})=45^\circ$, the feet(?) of the heights from the vertices $A,B,C$ are (in order) $D,E,F$, if the distance between $A$ and the line $EF$ is $6$, what is the perimeter of $DEF$=?
(Note:I am not well familiar with English Geometry terminology, as seen above)
My Attempts
Unfortunately I couldn't do so much this time, maybe due to something that I should know to be able to solve the question but don't know,
I have drawn triangle $DEF$...(1)
I have drawn three circles which took $A,B,C$ for centers and were tangent to $|DE|,|ED|,|EF|$....(2) (Thought it might work)
I have realised that I lacked some material about the usage of them...(3)
I have also started with the assumption that $AD \perp EF$ am I right about this?
What are your suggestions to proceed (or to start maybe, a more suitable word)?
Since $\Delta AEF\sim\Delta ABC$, we obtain $$\frac{AD}{6}=\frac{AB}{AE}=\frac{1}{\cos45^{\circ}}=\sqrt2.$$ Thus, $AD=6\sqrt2$ and $$P_{\Delta DEF}=\sum_{cyc}a\cos\alpha=\sum_{cyc}\frac{a(b^2+c^2-a^2)}{2bc}=$$ $$=\frac{1}{2abc}\sum_{cyc}a^2(b^2+c^2-a^2)=\frac{1}{2abc}\sum_{cyc}(2a^2b^2-a^4)=\frac{8S^2}{abc}=$$ $$=\frac{8\left(\frac{1}{2}bc\cos45^{\circ}\right)^2}{abc}=\frac{bc}{a}=\frac{h_a}{\sin45^{\circ}}\cdot\frac{\frac{1}{2}bc\sin45^{\circ}}{\frac{1}{2}ah_a}=\sqrt2AD=12.$$