MathLabs

Problem 4

Let ABCDABCD be a cyclic quadrilateral. Let P,Q,RP,Q,R be the feet of perpendiculars from DD to lines BC,CA,ABBC,CA,AB, respectively. Show that PQ=QRPQ=QR if and only if the bisectors of angles ABCABC and ADCADC meet on segment ACAC.
Step 7 of 7: Complete both implications
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In plain words

The projective midpoint criterion and the angle-bisector ratio criterion are the same condition.

PQ=QR  ⟺  cr⁡(P,R;Q,I∞)=−1  ⟺  ADCD=ABBC  ⟺  X=YPQ=QR\iff\operatorname{cr}(P,R;Q,I_\infty)=-1\iff\frac{AD}{CD}=\frac{AB}{BC}\iff X=Y
Detailed analysis

Steps 2 and 4 give PQ=QR  ⟺  cr⁡(P,R;Q,I∞)=−1PQ=QR\iff\operatorname{cr}(P,R;Q,I_\infty)=-1 and identify this with the harmonic pencil cr⁡(BC,BA;BF,BE)=−1\operatorname{cr}(BC,BA;BF,BE)=-1. Step 5 evaluates it as PQ=QR  ⟺  AD/CD=AB/BCPQ=QR\iff AD/CD=AB/BC, while Step 6 gives X=Y  ⟺  AD/CD=AB/BCX=Y\iff AD/CD=AB/BC. Therefore PQ=QRPQ=QR holds if and only if the two internal angle bisectors meet at the same point X=YX=Y on segment ACAC, as required.