ABC is a triangle. The angle bisector at A meets the side BC at X. The perpendicular to AX at X meets AB at Y. The perpendicular to AB at Y meets the ray AX at R. XY meets the median from A at S. Prove that RS is perpendicular to BC.
Solution
Let the line through C parallel to AX meet the ray BA at C'. Let the perpendicular from B meet the ray C'C at T and the ray AX at U. Let the line from C parallel to BT meet BA at V and let the perpendicular from V meet BT at W. So CVWT is a rectangle.
AU bisects ∠CAV and CV is perpendicular to AU, so U is the midpoint of WT. Hence the intersection N of AU and CW is the center of the rectangle and, in particular, the midpoint of CW. Let M be the midpoint of BC. Then since M, N are the midpoints of the sides CB and CW of the triangle CBW, MN = BW/2.
Since CC' is parallel to AX, ∠CC'A = ∠BAX = ∠CAX = ∠C'CA, so AC' = AC. Let A' be the midpoint of CC'. Then AU = C'T - C'A'. But N is the center of the rectangle CTWV, so NU = CT/2 and AN = AU - NU = C'T - C'A' - CT/2 = C'T/2. Hence MN/AN = BW/C'T. But MN is parallel to BW and XY, so SX/AX = MN/AN = BW/C'T.
Now AX is parallel to VW and XY is parallel to BW, so AXY and VWB are similar and AX/XY = VW/BW = CT/BW. Hence SX/XY = (SX/AX) (AX/XY) = CT/C'T.
YX is an altitude of the right-angled triangle AXR, so AXY and YXR are similar. Hence XY/XR = XA/XY. But AXY and C'TB are similar, so XA/XY = C'T/BT. Hence SX/XR = (SX/XY) (XY/XR) = (CT/C'T) (C'T/BT) = CT/BT. But angles CTB and SXR are both right angles, so SXR and CTB are similar. But XR is perpendicular to BT, so SR is perpendicular to BC.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002