Problem:
Let ABCD be a cyclic quadrilateral. Prove that there exists a point X on segment BD such that ∠BAC=∠XAD and ∠BCA=∠XCD if and only if there exists a point Y on segment AC such that ∠CBD=∠YBA and ∠CDB=∠YDA.
Solution:
By the symmetry, it suffices to show the "only if" part by assuming that there exists a point X on segment BD such that ∠BAC=∠XAD and ∠BCA=∠XCD.
Because ABCD is cyclic, we have ∠XAD=∠BAC=∠BDC=∠XDC and ∠XDA= ∠BDA=∠BCA=∠XCD. Hence triangles AXD and DXC (and ABC ) are similar to each other. In particular,
DXAX​=XCDX​ or DX2=AX⋅CX
Because ∠BAC=∠XAD, we have ∠BAX=∠CAD. Because ABCD is cyclic, we have ∠CAD=∠CBD=∠CBX. Consequently, ∠BAX=∠CBX. Note that
∠AXB=∠XAD+∠ADX=∠BAC+∠ACB=∠BDC+∠DCX=∠CXB
From the above facts, we conclude that triangles ABX and BCX (and ACD ) are similar to each other and so we have BX2=AXâ‹…CX. Thus, BX2=AXâ‹…CX=DX2; that is, X is the midpoint of the segment BD. Therefore
BCAB​=XCDX​=XCBX​=DCAD​ or CDBC​=ADBA​
Construct point Y on segment AC such that ∠CBD=∠YBA. From ∠CBD=∠YBA and ∠BAY=∠BAC=∠BDC, we conclude that triangles BAY and BDC are similar to each other, from which it follow that
YABY​=CDBC​=ADBA​ or BABY​=ADAY​
Note also that ∠YBA=∠CBD=∠CAD=∠YAD. We conclude that triangles BYA and AYD are similar to each other, implying that ∠CDB=∠YAB=∠YDA. This is the desired point Y.
OR
By symmetry, it suffices to show that there exists X on the segment BD such that ∠BAC=∠XAD and ∠BCA=∠XCD if and only if AB⋅CD=AD⋅BC.
There is a unique point X1​ on segment BD such that ∠X1​AD=∠BAC. There is a unique point X2​ on segment BD such that ∠BCA=∠X2​CD. Because ABCD is cyclic, ∠BCA=∠BDA=∠X1​DA. Hence triangles ABC and AX1​D are similar to each other, implying that
BCAC​=X1​DAD​
Likewise, we can show that ABC and DX2​C are similar to each other and ACAB​=DCDX2​​. Multiplying the last two equations together gives
BCAB​=ACAB​⋅BCAC​=DCDX2​​⋅X1​DAD​
from which it follows that
AD⋅BCAB⋅CD​=DX1​DX2​​
Note that point X exists if and only if X1​=X2​, or DX2​=DX1​; that is, AB⋅CD= AD⋅BC.
The problems on this page are the property of the MAA's American Mathematics Competitions