Problem:
Let ABCD be a convex quadrilateral with AB=2,AD=7, and CD=3 such that the bisectors of acute angles β DAB and β ADC intersect at the midpoint of BC. Find the square of the area of ABCD.
Solution:
Let I be the midpoint of BC, and let R and S denote the reflections of C and B across DI and AI, respectively. Note that, because DI and AI are angle bisectors, R and S lie on segment AD. Then AS=2 and DR=3, so RS=2. Furthermore, IR=IC=IB=IS, so
β ABI=β ASI=β DRI=β DCI
and, if T is the foot of the perpendicular from I to AD, then RT=ST=1.
Now let Ξ±=β BAI=β IAD,Ξ²=β CDI=β IDA, and Ξ³=β ABC=β BCD. Because the sum of the measures of the interior angles of a quadrilateral is 360β, it follows that Ξ±+Ξ²+Ξ³=180β. This implies
β BIA=180ββΞ³βΞ±=Ξ²
and likewise β CID=Ξ±. Thus β³ABIβΌβ³ICD, so BI=CI=ABβ CDβ=6β.\
Finally, applying the Pythagorean Theorem to β³IRT yields IT=5β, so
Let X be the intersection of rays AB and DC. Then the angle bisectors of β DAB and β CDA meet at the incenter I of β³XAD. Because XI is both an angle bisector and a median of β³BXC, it is the perpendicular bisector of BC. Let T,E, and F be the projections of I onto lines AD,XA, and XD, respectively.
Because $\triangle B X I$ is congruent to $\triangle C X I$, it follows that $C F=B E$. By equal tangents, $D T=D F=3+C F=$ $3+B E=1+A E=1+A T$. Because $D T+A T=A D=7, A T=3$ and $D T=4$ from which $C F=B E=$ $A E-A B=A T-A B=3-2=1$. Let $r$ be the inradius of $\triangle A D X$. Because $\triangle B E I \sim \triangle I E X$, it follows that $X E=\frac{I E^{2}}{B E}=r^{2}$.\\
Because the inradius of a triangle with sides $a, b$, and $c$ and semiperimeter $s$ is
