In standard orthic configuration , find AB + AC.

Question

Consider standard orthic configuration notation.

$\angle FA'E=30$
$\angle BAD : \angle CAD = 2 : 3$ . Given $~BD$ = 3 , find $~AB + AC $.

Answer 1

Since $\angle BFC = \angle BEC = 90^{\circ}$, $BFEC$ is cyclic and the center of their circle is $A'$. Thus, from $\angle FA'E = 30^{\circ}$, $\angle FBE = \angle FCE = 15^{\circ}$ follows.

If $\angle BAD = 2\varphi$ and $\angle CAD = 3\varphi$, then $\angle BAC = 5\varphi$. From the right triangle $BAE$ $5\alpha = 75^{\circ}$ and $\alpha=15^{\circ}$. So $\angle BAD = 30^{\circ}$, $\angle CAD = 45^{\circ}$.

Now from $BD=3$ $AB=6$ follows. Thus $AD=3\sqrt{3}$ and $AC=3\sqrt{6}$.