Dokaži, da je Euclidova desna stopala Teorema 1 in 2: ET_1 => navzgor {BC} ^ {2} = navzgor {AC} * na sliki {CH}; ET'_1 => bar (AB) ^ {2} = bar (AC) * bar (AH); ET_2 => barAH ^ {2} = navzgor {AH} * levo {CH}? [tukaj vnesite vir slike] (https

Odgovor:

Glej dokaz v oddelku o razlagi.

Pojasnilo:

Opazimo to, v #Delta ABC in Delta BHC #, imamo, # / _B = / _ BHC = 90 ^ @, "common" / _C = "common" / _BCH in,., #

# / _A = / _ HBC rArr Delta ABC "je podoben" Delta BHC # #

Njihove ustrezne strani so sorazmerne.

#:. (AC) / (BC) = (AB) / (BH) = (BC) / (CH), t.j. (AC) / (BC) = (BC) / (CH) #

#rArr BC ^ 2 = AC * CH #

To dokazuje # ET_1 #. Dokaz. T # ET'_1 # je podobna.

Dokazati # ET_2 #, to pokažemo #Delta AHB in Delta BHC # so

podobno.

V #Delta AHB, / _AHB = 90 ^ @:. /_ABH+/_BAH=90^@……(1)#.

Tudi, # / _ ABC = 90 ^ @ rArr /_ABH+/_HBC=90^@………(2)#.

Primerjava # (1) in (2), /_BAH=/_HBC…………….(3)#.

Tako, v #Delta AHB in Delta BHC, # imamo, # / _ AHB = / _ BHC = 90 ^ @, /_BAH=/_HBC…………. ker, (3) #

#rArr Delta AHB "je podoben" Delta BHC. #

#rArr (AB) / (BC) = (BH) / (CH) = (AH) / (BH) #

Iz # 2 ^ (nd) in 3 ^ (rd) "razmerje", BH ^ 2 = AH * CH #.

To dokazuje # ET_2 #