Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 100Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry, TutoringPost your solutions or ideas in the comments.
This follows quite trivially from the fact that AC²+BC²+2CD²=AD²+BD²=AB².
name x radius for AC, y for CB, z for CD, R for ABname Sa area above 1 & 2, Sb left of 1, Sc right of 2S1 + Sb = x²/2 (1)S2 + Sc = y²/2 (2)S1 + S2 + S3 + Sa = x∙y (3)( from euklid theorem (2z)²=2x∙2y => z² = x∙y )S1 + S2 + S3 + S4 + S5 + Sa + Sb + Sc = R²/2 + S3(S1+Sb)+(S2+Sc)+(S1+S2+S3+Sa)-S1-S2+S4+S5 = R²/2 + S3substitute (1), (2), (3)(x²/2 + y²/2 + x∙y) - S1 - S2 + S4 + S5 = R²/2 + S3R²/2 - S1 - S2 + S4 + S5 = R²/2 + S3 (R=x+y, R²=x²+2xy+y², R²/2 = x²/2 + xy + y²/2)S4 + S5 = S1 + S2 + S3
Let S(AB), S(AC), S(BC) and S(CD) be the areas of the semicircles with diameters AB, AC, BC and CD, respectively.We have S(AB) = 2S(CD) – S3 + S4 + S5 ++ S(AC) – S1 + S(BC) – S2, soS1 + S2 + S3 = S4 + S5 + S(AC) + S(BC) ++ 2S(CD) – S(AB).Thus, to prove that S1 + S2 + S3 = S4 + S5, it’s enough to prove thatS(AB) = S(AC) + S(BC) + 2S(CD).We have S(AC) + S(BC) + 2S(CD) == (pi/2)[AC^2/4 + BC^2/4 + 2.CD^2/4] == (pi/2)[AC^2/4 + CD^2/4 + BC^2/4 + CD*2/4] == (pi/2)[(AD^2/4 + BD^2/4] = (pi/2)(AB^2/4) == S(AB).Q.E.D.
let the midpoint of AB is O, connect ODlet "L","M","N"and"R" be the names of circles having the diameter AB,CB,AC and CD respectivelysay CB/2=x & AC/2=yso , radius of "L"=2(x+Y)/2=x+yapply Pythagoras theorem in triangle DCOso the radius of "R"=2√xy/2 = √xyArea of L=1/2∏(x+y)^2area of M=1/2∏x^2area of N=1/2∏y^2L-M-N = ∏xyarea of R= ∏xyso area of R= area of L-M-NR- (R-(S1+S2+S3)) = (L-M-N)- (R-(S1+S2+S3))so,S1+S2+S3=S4+S5