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
The square of the area of ABCD is (65β)2=180β.
OR
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 β³BXI is congruent to β³CXI, it follows that CF=BE. By equal tangents, DT=DF=3+CF=3+BE=1+AE=1+AT. Because DT+AT=AD=7, AT=3 and DT=4 from which CF=BE=AEβAB=ATβAB=3β2=1. Let r be the inradius of β³ADX. Because β³BEIβΌβ³IEX, it follows that XE=BEIE2β=r2.
Because the inradius of a triangle with sides a,b, and c and semiperimeter s is
