Proposed Problem
Click the figure below to see the complete problem 413 about Cyclic Quadrilateral, Orthocenter of a triangle, Parallelogram, Concurrency, Congruence, Altitude.
See also:
Complete Problem 413
Level: High School, SAT Prep, College geometry
Monday, January 4, 2010
Problem 413: Cyclic Quadrilateral, Orthocenter, Parallelogram, Concurrency, Congruence
Subscribe to:
Post Comments (Atom)
Can you show that, in general, two quadrilaterals HFEG and ABCD have the same area?
ReplyDelete1. BCFE is a parallelogram
ReplyDeleteBE and CF cut circle (ABCD) at E’ and F’ .
Since E is the Orthocenter of triangle ABD , so E’ is the symmetric point of E over AD
Similarly F’ is the symmetric point of F over AD
BE//CF
BCE’F’ and BE’F’C are isosceles trapezoids and angle(BCF)=angle(CF’E’)=angle(EFF’)
So BC//EF and BCFE is a parallelogram
AGHD is a parallelogram
We have AG // HD ( both line perpen. to BC )
AG and HD cut the circle at G’ and H’ .
With the same logic as above we have GG’H’H and DH’G’A are isosceles trapezoids and GH //AD
So AGHD is a parallelogram
In the same way ABHF and CDEG are the parallelograms
2. Diagonals of a parallelogram bisect each other at mid point . Appling this propertie in above 4 parallelograms we will get the result.
3. Opposite sides of a parallelogram are congruence and opposite angles of a parallelogram are congruence.
( properties of parallelogram) . Applying these properties in above 4 parallelograms we have ABCD and HFEG have 4 sides and 4 internal angles congruence to each other . So they are congruence
Peter Tran
Problem 413
ReplyDeleteSince the problem 408 follows that BC=//EF , GE=//CD, GH=//AD,AB=//HF so BCFE, BHFA, AGHD ,ABHF are parallelograms.The (AH,BF),(GD,EC),(BF,EC),(BF,AH) intersecting at their mid. Which is the same point.
Triangle ABC=triangle HFE and triangle ACD=triangle HEG so quadr.ABCD=quadr.EFHG.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE