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 5 of 7: Evaluate the projected condition metrically
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In plain words

The two right-angle circles turn the cross-ratio condition into a ratio of opposite sides.

PQ=CDsin⁡∠BCA,QR=ADsin⁡∠BAC,PQ=QR  ⟺  ADCD=ABBCPQ=CD\sin\angle BCA,\qquad QR=AD\sin\angle BAC,\qquad PQ=QR\iff\frac{AD}{CD}=\frac{AB}{BC}
Detailed analysis

The cyclic quadrilateral CQDPCQDP has diameter CDCD, so its chord formula gives PQ=CDsin⁡∠BCAPQ=CD\sin\angle BCA. Likewise AQDRAQDR has diameter ADAD, giving QR=ADsin⁡∠BACQR=AD\sin\angle BAC. The sine rule in ABCABC gives sin⁡∠BCA/sin⁡∠BAC=AB/BC\sin\angle BCA/\sin\angle BAC=AB/BC. Therefore PQ=QR  ⟺  CD AB/BC=AD  ⟺  AD/CD=AB/BCPQ=QR\iff CD\,AB/BC=AD\iff AD/CD=AB/BC. This is the explicit metric evaluation of the harmonic pencil in Step 4.