Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view the complete problem 994
Friday, March 21, 2014
Geometry Problem 994: Trapezoid, Midpoints of the bases, Concurrent Lines
Subscribe to:
Post Comments (Atom)
Let CD cut MN at P
ReplyDelete∆ PNC similar to ∆PMD
So PN/PM=NC/MD=NB/MA => ∆ PNB similar to ∆ PMA
So A, B, P are collinear and AB, MN, DC concurrent at P
Problem 994
ReplyDeleteIs BN/AM=NC/MD then from the inverse theorem batch AB,MN and DC are concurrent at P.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
M is the middle AD, P is the intersection of AB and CD, N is the intersection of BC and MP
ReplyDeleteWe have to demonstrate that N is the middle of BC
AD//BC => ∠MAP=∠NBP and ∠MDP=∠NCP
=> Δ MAP is similar to Δ NBP (AA)
=> (1) PN/NB = PM/MA
=> Δ MDP is similar to Δ NCP (AA)
=> (2) PM/MD = PN/NC
MA=MD => (3) PM/MA = PM/MD
(1&3&2) PN/NB = PM/MA = PM/MD = PN/NC
=> PN/NB= PN/NC => NB=NC
=> N is the middle of BC