On the equal AC and BC of an isosceles right triangle ABC , points D and E are marked respectively, so that CD=CE. Perpendiculars on the straight line AE, passing through the points C and D, intersect the side AB at the points P and Q.Prove that BP=PQ.
Problem
Source: Tuymaada Junior 2006 p1
Tags: geometry, right triangle, isosceles, equal segments
06.12.2021 17:08
Another nice solution at https://stanfulger.blogspot.com/2021/12/tuymaada-junior-2006.html Best regards, sunken rock
09.12.2021 01:28
Let D be on ¯AC and let E be on ¯BC with ¯BC being a horizontal line parallel to the x-axis. We proceed with coordinate bashing. Let point C be the origin. We then let B=(a,0), A=(0,a), E=(b,0), and D=(0,b). We then proceed by finding the equations of a few lines. ¯AB−y=−x+a¯AE−y=−abx+aSince ¯CP and ¯DQ are perpendicular to ¯AE, they both have a slope of ba. We then use points D and C to find the equations of these two lines. ¯PC−y=bax¯DQ−y=bax+bNext, we find that the point of intersection of ¯AB and ¯PC (which is point P) is (a2a+b,aba+b). We then find that the intersection of ¯AB and ¯DQ (which is point Q) is (a2−aba+b,2aba+b). The final thing to do is find the distances BP and PQ (we will multiply the lengths by (a+b) for simplicity). (a+b)(BP)=√(a2+ab−a2)2+(0−ab)2=√2a2b2(a+b)(QP)=√(a2−a2+ab)2+(ab−2ab)2=√2a2b2Since BP=QP, our proof is complete. ◻