AoPS Academy
be a cyclic quadrilateral and let the intersection point of lines 
and 
be 
. Let the points 
and 
be arbitrary points on sides and respectively, which satisfy the conditions
Prove that 
.
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray 
at points 
and , respectively. Given that 
, prove that 
.
of triangle 
meet its circumcircle 
at point 
. The circle 
with diameter 
(
is the orthocenter of ) meets for the second time at point 
. Restore the triangle if the points , , are given.
using only ruler and compass, then yes., call it 
, now we draw 
, also using compass we can also draw the reflection of a point over a segment clearly, so we can draw 
which is reflection of 
over 
, which in fact by isogonality of 
and the A-altitude in we get that 
, now we draw 
, from spiral sim propeties of which is the A-queque point, we know that we just drawn the orthocenter, so now we draw the perpendicular bisector of 
which is clearly line and since we can draw we are done, thus we can restore and we finished 
of triangle meet its circumcircle at point . The circle with diameter ( is the orthocenter of ) meets for the second time at point . Restore the triangle if the points , , are given.
; Furthermore, 
cuts at its midpoint. Taking this into account, the construction of triangle , given 
, is immediate:
The center of the circle (
) is constructed, which must be the circumcenter of . The perpendicular to 
through cuts at 
(midpoint of ). The points of intersection of the circle () with the perpendicular to through are the vertices 
and 
of the triangle to be constructed. and are fixed, the construction of triangle is NOT possible if is located in the region between the perpendiculars to in and .
with center then draw a line through parallel to 
it intersects perp.bisector of at midpoint of so is constructible. next cuts at 
antipode of 
then take midpoint of 
also midpoint of 
then just take line through perpendicular to 
to get 
and 
.
through 
,