Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 884.
http://img442.imageshack.us/img442/5306/problem8841.pngPartial answer to the problem.Questions 2, 5, 6 . note that ∆RCQ and ∆MXQ are congruent ( case SAS)So MQ=QR and ∠ (CQR)= ∠ (MQX)=yIn isosceles triangle MQH we have ∠MQH)+2x=180At point Q we have ∠ (MQH)+2y+180=360 => ∠ (MQH)+2y=180So x=y and MH//XQ//ACSimilarly DE//PY//BALet Q’ is the projection of Q over QC and P’ is the projection of P over NBWe have QQ’=CQ.sin(C )=b.c/a and PP’=PB.sin(B)=c.b/aSo QQ’=PP’Note that QQ’=GG’ and PP’=FF’ ( congruent triangles)So FG//NR//BCQuestions 9, 12, 13Note that right triangles XAY, MX’X and YY’D are congruent ( case HA)So MX’=AX= b => MH=4bSo A’X’AY’ is a square and AA’ is a diagonal so AA’ is a bisector of angle A and A’ . AA’ ⊥ PQ