From Gogeometry Using : Additional points congruent triangles Solution of a previous problem 218 DB=GF+FJ From problem 217, ΔJBF is congruent to ΔHDE (AAA) => JF=EH Therefore a+b=h
From Gogeometry Using : Pythagoras Inscribed angles in a circle central and inscribed angle Concyclic points Solution of a previous problem : Pb367 AD⊥AH => A, D, H are concyclic Arc EH : ∠EDH=45° and ∠EAH=45° => A, D, H and E are concyclic Arc AE : ∠EDA=45° => ∠EHA=45° […]
From Gogeometry Using : Pythagoras Define a,b and x respectively hypothenuse of S1, S2 and S S1 is a square => S1= a^2/2 In the same way, S2= b^2/2 and S= x^2/2 From Pb 367 we know that x^2=a^2+b^2 => (x^2)/2=(a^2)/2+(b^2)/2 Therefore S=S1+S2
From Gogeometry Using : Isometric transformation Additional points Pythagoras congruent triangles Define H’ rotation of H with angle –90° and center A =>AH=AH’ and BH’=DH=b => BH’ ⊥DH <=> BH’ ⊥BG ∠HAG=45° and ∠HAH’=90° => ∠GAH’=45° ΔGAH is congruent to ΔGAH’ (SAS) => GH’=GH=x BH’=b, BG=a, BH’ ⊥BG and GH’=x […]
From Gogeometry Using : Additional points Area of a triangle Define a = side of square ABCD Define h1,h2,h3 and h4 the heights respectively of S1, S2, S3 and S4 h1+h3=a, h4-h2=a S=[ABCD]=a^2 2S1=a.h1, 2S3=a.h3 2S1+2S3=a(h1+h3)=a^2 2S2=a.h2, 2S4=a.h4 2S4-2S2=a(h4-h2)=a^2 Therefore <b>S1+S3=S4-S2=S/2</b>
From Gogeometry Using : Area of a triangle Define h1,h2,h3 and h4 the heights respectively of S1, S2, S3 and S4 Define a the side of the square ABCD S1=a.h1/2 S2=a.h2/2 S3=a.h3/2 S4=a.h4/2 And h1+h3=h2+h4=a S1+S3=a(h1+h3)/2=a^2/2 S2+S4=a(h2+h4)/2=a^2/2 S=[ABCD]=a^2 Therefore S1+S3= S2+S4=S/2
From Gogeometry Using : Isometric transformation Additional points congruent triangles Area of a triangle Define a square ABCD Define E in AB, AE=a and EB=b Define F in BC such as BF=b and FC=a [ABCD]=[HIGD]+[EBFI]+[AEIH]+[FCGI] [AEIH]=[FCGI]=ab=>[ABCD]=a^2+b^2+2ab=(a+b)^2 Cut [AEIH] in 2 triangles T1=AEI and T2=AHI Cut [FCGI] in 2 triangles T3=FCG […]
From Gogeometry Using : Pythagoras Similar triangles In ΔBHC : ∠ABC + ∠HCB =90° In ΔBCA : ∠HCA + ∠HCB =90° => ∠HCA = ∠ABC=β In ΔCHA : ∠HCA +∠BAC=90° => ∠HCB =∠BAC= α => ΔBHC is similar to ΔCHA (aaa=β90α) and ΔBHC is similar to ΔBCA (aaa=β90α) Therefore ΔBHC, […]