Proposed Problem
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Complete Problem 292
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Saturday, May 23, 2009
Problem 292: Triangle, Parallel, Congruence, Similarity
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suppose BB' meet to CC' at P1 and AA' meet to BB' at P2
ReplyDelete=> AA' meet to CC' at P3
from thales theorem
(P1P2+P1B')/BB' = P2A'/AA' ( A'B'//AB ) (1)
(P2P3+P2A')/AA' = P3C'/CC' ( A'C'//AC ) (2)
(P1P3+P3C')/CC' = P1B'/BB' ( C'B'//CB ) (3)
substitute (3) & (2) at (1)
P1P2/AA' + P1P3/CC' + P3C'/CC' = P3C'/CC' - P2P3/AA'
P1P2/AA' + P1P3/CC' = - P2P3/AA'
the sum of two positive nr must be positive
( the length are always positive )
contradiction came from our wrong supposition
=>
P1=P2=P3=P
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name K, GP meet DD', name L, PN meet FF', name T, PM meet EE'
▲ADA' ~ ▲A'KP => AA'/PA' = m/KP (1)
▲FCC' ~ ▲PLC' => CC'/C'P = m/PL (2)
▲PA'C' ~ ▲PAC => PA'/AA' = PC'/CC' (3)
from (1) & (2) & (3)
m/KP = m/PL => KP = PL =>
PG = m + KP
PN = m + PL
=>
PG = PN = d
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extend MP to R ( R on AC), extend NP to S ( S on AB)
▲RCM
RN/RP = NC/PM => RN/a = c/d => RN = (ac)/d (1)
▲ANS
RN/PN = AR/PS => RN/d = d/b => RN = d²/b (2)
from (1) & (2)
(ac)/d = d²/b =>
d³ = abc
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All facts are supported
Thales theorem and paralellograms formed by paralell lines