# Paucellier vs. Hart's A-frame

Author:
asifsound
■ Very interesting in above Fig, AB×BD cos(∠ABD) has 2 results. ① AB×BH = variable × variable ---- = long × short ② AB cos(180ﾟ- ∠ABO)×BD = BO×BD = constant × constant [ ① is the same ②. ] Further, AB2 = AD2 + BD2 - 2 AD×BD cos(∠ADB) so, AD2 - AB2 = -BD2 + 2 AD×BD cos(∠ADB) Here, AD×BD cos(∠ADB) = DA×DH' = variable × variable ---- ③ [where, DH' = BD cos(∠ADB) ], or = AD cos(∠ADB)×BD = DO×BD = constant × constant ---- ④　　[ ③ is the same ④. ] Then, Gap of ① and ③: ④ - ② = DO×BD - BO×BD = (DO-BO)×BD = BD2 = constant ---------- | ◆ More easy recognition: | △DBH ∽ △ABO ---so, BH/DB = BO/AB, so, BH×AB = BO×DB --- ①② | △ADO ∽ △BDH' ---so, DO/AD = DH'/BD, so, DO×BD = DH'×AD --- ③④ ■ Trick In above under Figure, distance ( from middle [= MM] to vertical [= HH]) is orange MMHH . This is the same 0.5 AACC - HHCC = 0.5 (AACC - 2 HHCC) So, 0.5 (AACC) × 0.5 (AACC -2 HHCC) = 0.25 AACC × AAC'C' [where, C'C' is mirror point CC against HH] |--------| AAC'C' |------------| AACC --- this is orthodox Paucellier inversor result. [AAMM×MMHH = MMCC×MMHH , is more honest.] ■ Hart's Inversor butterfly SMMTHH is Hart's inversor. ST // MMHH, and, ST × MMHH = constant, So, horizontal line between S and MM, ---- ex. SS' = q SMM (; inner divided point, q = inner ratio) S' M' H' T' |---|----|---| any horizontal segment (S' is the cross node with SMM segment, node M'/H'/T' are ditto.) ST × MMHH = constant --- So, S'M' × M'T' = constant. --- ∵ q(1-q) = constant, too. and S'M' = H'T' so, S'M' × S'H' = constant. Power of a point OA = aa, OB = aa, OC = bb, H is middle of AB, and OH ⊥ AB, A, C, H, B are colinear (= on the same line). if so, AC×CB = constant Proof: OB2 = HB2 + OH2 OC2 = HC2 + OH2 so, OB2 - OC2 = HB2 - HC2 = aa2 - bb2 = constant and, HB2 - HC2 = (HB+HC)×(HB-HC) = BC×AC Then AC×CB = constant