Given acute triangle $\triangle ABC$ satisfying $|\overline{AB}| \ne |\overline{AC}|$. Let $D,E$, respectively, be the midpoints of $\overline{AB}, \overline {AC}$. Let $Q, P$ be the intersections of $(\triangle ADE)$ and $(\triangle BCD)$, $(\triangle ADE)$ and $(\triangle BCE)$, respectively. Prove that $|\overline{AP}| = |\overline{AQ}|$.
I have already tried using radical axis but still cannot figure out the solution. Please help me with this. Thanks.
Best Answer
Draw a line through the midpoint between BD perpendicular to BD. Draw a line through the midpoint of CE perpendicular to CE. Let the point where these meet be labeled K. Draw the line AK. From here it should be rather straightforward that H, the center of the circle circumscribing ADE, lies on AK and that AK bisects the angle GAF. Since the triangle AGF is also circumscribed, AG = AF.
EDIT: Further explanations in the figure below. Draw two lines, not parallel to each other, of which AC and AB are segments, with A being their point of intersection. Draw any two parallel lines crossing the other two, neither of which pass through A. In the figure they intersect at C, B, D, and E. Draw perpendicular lines through the midpoints of the segments CD, AD, BE, AE. Let the points where these meet be G and F. Now G, F, and A are colinear.
Incidentally, the same steps are perfomed when circumscribing the circles above. Hence, A, H, and K are colinear in the figure above.
EDIT: Now, draw the circle with center K and radius KF, as in the figure below. Note the intersections F and G. From this it should be obvious that KA bisects the angle GAF.