tag:blogger.com,1999:blog-6933544261975483399.post8426144364607133074..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 994: Trapezoid, Midpoints of the bases, Concurrent LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-55192360188057104632023-12-14T14:05:27.400-08:002023-12-14T14:05:27.400-08:00M is the middle AD, P is the intersection of AB an...M is the middle AD, P is the intersection of AB and CD, N is the intersection of BC and MP<br />We have to demonstrate that N is the middle of BC<br />AD//BC => ∠MAP=∠NBP and ∠MDP=∠NCP<br />=> Δ MAP is similar to Δ NBP (AA)<br />=> (1) PN/NB = PM/MA<br />=> Δ MDP is similar to Δ NCP (AA)<br />=> (2) PM/MD = PN/NC<br />MA=MD => (3) PM/MA = PM/MD<br />(1&3&2) PN/NB = PM/MA = PM/MD = PN/NC<br />=> PN/NB= PN/NC => NB=NC<br />=> N is the middle of BC<br />rv.littlemanhttps://www.blogger.com/profile/05572092955468280791noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18378923413410070342016-08-31T00:50:46.335-07:002016-08-31T00:50:46.335-07:00Problem 994
Is BN/AM=NC/MD then from the inverse...Problem 994<br />Is BN/AM=NC/MD then from the inverse theorem batch AB,MN and DC are concurrent at P.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38952514599775675562014-03-21T10:17:46.477-07:002014-03-21T10:17:46.477-07:00Let CD cut MN at P
∆ PNC similar to ∆PMD
So PN/PM=...Let CD cut MN at P<br />∆ PNC similar to ∆PMD<br />So PN/PM=NC/MD=NB/MA => ∆ PNB similar to ∆ PMA<br />So A, B, P are collinear and AB, MN, DC concurrent at P<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com