tag:blogger.com,1999:blog-6933544261975483399.post8076750670538569384..comments2024-06-15T03:59:07.072-07:00Comments on Go Geometry (Problem Solutions): Problem 413: Cyclic Quadrilateral, Orthocenter, Parallelogram, Concurrency, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-38054133488343586232016-06-17T14:49:34.295-07:002016-06-17T14:49:34.295-07:00Problem 413
Since the problem 408 follows that BC=...Problem 413<br />Since 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.<br />Triangle ABC=triangle HFE and triangle ACD=triangle HEG so quadr.ABCD=quadr.EFHG.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br /><br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13288533728371302472011-03-31T01:14:41.484-07:002011-03-31T01:14:41.484-07:001. BCFE is a parallelogram
BE and CF cut circle (...1. BCFE is a parallelogram<br /> BE and CF cut circle (ABCD) at E’ and F’ .<br />Since E is the Orthocenter of triangle ABD , so E’ is the symmetric point of E over AD <br />Similarly F’ is the symmetric point of F over AD <br />BE//CF<br />BCE’F’ and BE’F’C are isosceles trapezoids and angle(BCF)=angle(CF’E’)=angle(EFF’)<br />So BC//EF and BCFE is a parallelogram <br />AGHD is a parallelogram<br />We have AG // HD ( both line perpen. to BC )<br />AG and HD cut the circle at G’ and H’ .<br />With the same logic as above we have GG’H’H and DH’G’A are isosceles trapezoids and GH //AD <br />So AGHD is a parallelogram<br />In the same way ABHF and CDEG are the parallelograms<br /><br />2. Diagonals of a parallelogram bisect each other at mid point . Appling this propertie in above 4 parallelograms we will get the result.<br /><br />3. Opposite sides of a parallelogram are congruence and opposite angles of a parallelogram are congruence. <br />( 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<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78931676615057713912010-01-06T17:47:37.232-08:002010-01-06T17:47:37.232-08:00Can you show that, in general, two quadrilaterals ...Can you show that, in general, two quadrilaterals HFEG and ABCD have the same area?Anonymousnoreply@blogger.com