[/quote]
,
be an acute triangle. Let 
be a point on side 
and 
be a point on side 
such that lines 
and 
are parallel. Let 
be an interior point of 
.
and 
meet side 
and 
,
and 
.
and 
intersect at a point 
.
,
are collinear.
.
and 
,
as the greatest exponent 
such that 
divides 
.
and infinitely many 
such that 
.
and 
.
is given. Let 
to 
be the intersection point of 
lies on the arc 
.
.
be a trapezium inscribed in a circle 
with diameter 
. The circle with center 
meets 
and 
(where 
at 
.
is perpendicular to 
.
![\[ x + y^2 = p^m \]](http://latex.artofproblemsolving.com/e/c/f/ecf553036f22327dd457dc06611b6bc1b0ef072c.png)
![\[ x^2 + y = p^n \]](http://latex.artofproblemsolving.com/5/b/d/5bd672934a395aabd6865ca4b50d8da75c0fe07e.png)
For 
positive integers and 
being a prime.
and 
.
Equality holds when 
or 
.
at 
intersect again at 
.
Equality holds when 
intersect its circumcircle 
.
determine a triangle 
,
is relative to 
are concurrent at a point 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
intersect 
at 
and 
are either parallel or coincide.
and 
.
Where 
Y by Adventure10, Mango247
1.Let ABC be an isosceles triangle with AB=AC . Point D and P lie on the sides of AB and AC
,respectively . The line passing through B and parallel to AC meets the line DE at F . The line
passing through C and parallel to AB meets the line DE at G . Prove that Area[DBCG]/Area[FBCE]=AD/AE
2. Let ABCD be a cyclic quadrilateral and let P and Q be points on the sides AB and AD respectively such
that AP=CD and AQ=BC let M be the point of intersection of and PQ . Show that M is the midpoint of PQ
Thankyou
,respectively . The line passing through B and parallel to AC meets the line DE at F . The line
passing through C and parallel to AB meets the line DE at G . Prove that Area[DBCG]/Area[FBCE]=AD/AE
2. Let ABCD be a cyclic quadrilateral and let P and Q be points on the sides AB and AD respectively such
that AP=CD and AQ=BC let M be the point of intersection of and PQ . Show that M is the midpoint of PQ
Thankyou
be the heights from B and C in the triangle ABC. Then 
![$ Area[DBCG]=\frac{(DB+CG)h_c}{2}, \ Area[FBCE]=\frac{(BF+CE)h_c}{2}$](http://latex.artofproblemsolving.com/8/e/0/8e07638f419e9ad319af334df70743d938484817.png)
![$ \frac{Area[DBCG]}{Area[FBCE]}=\frac{DB+CG}{BF+CE}$](http://latex.artofproblemsolving.com/9/5/2/95293da68b6aa051795cfea438bc1f152d5d96d6.png)
and 
.
is a parallelogram, so 
bisects 
.
,
so 
.
