From The Mathematics Student Journal
ACJD, CBGH and BAEF are squares constructed outwardly on the sides of triangle ABC. DE, FG and HJ are drawn. If the sum of the areas of squares BAEF and CBGH is equal to the area of the rest of the figure, find the measure of angle ABC.
My Solution:
Let [ABC] denotes the area of triangle ABC and angle ABC = α, angle BAC = β, angle BCA = γ. Given that the sum of [BAEF] and[CBGH] is equal to area of the rest of the figure, we have:
(AB)2 + (BC)2 = [FBG] + [ABC] + (AC)2 + [AED] + [HCJ] (1)
Now [FBG] = [ABC] = [AED] = [HCJ] because
[FBG] = (1/2)(FB)(BG)sin(180 – α) = (1/2)(AB)(BC)sin α = [ABC] ,
[AED] = (1/2)(AE)(AD)sin(180 – β) = (1/2)(AB)(AC)sin β = [ABC]
and [HCJ] = (1/2)(HC)(CJ)sin(180 – γ) = (1/2)(BC)(AC)sin γ = [ABC] .
Equation (1) becomes
(AB)2 + (BC)2 = (AC)2 + 2(AB)(BC)sin α
Replacing (AC)2 using cosine law,
(AB)2 + (BC)2 = (AB)2 + (BC)2 – 2(AB)(BC)cos α + 2(AB)(BC)sin α
cos α = sin α
α = angle ABC = 45 degrees.
