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

Labels:
congruence,
parallel,
similarity,
triangle

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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