abcd is a square .from the diagonal bd, a length bx is cut off equal toba.from x, a straight line xy is drawn perpendicular to bd to meet ad at y.then ab+ay=
Call the side of the square S.....
Using the Law of Cosines, we have
AX^2 = 2S^2 -2S^2cos(45) = 2S^2 - 2S^2(1/√2) = S^2 [ 2 - √2]
So........AX = S*√[ 2 - √2]
And using some basic geometry <YXD = 90 and <XDA = 45......so <XYD = 45....so <AYX = [180- <XYD]= 135
And since AB = BX and <ABD = 45, then <AXB = (180 - 45]/2 = 67.5
Then <AXY = [180 - 90 - 67.5] = 22.5
And using the Law of Sines, again, we have
AY/sin(22.5) = AX/sin(135)
AY = sin(22.5)/sin(135)* S*√[ 2 - √2] = [√[1-1/√2] / √2] * √2* √[2 - √2]S = [√[1-1/√2]*√[2 - √2]*S = [ √[√2 -1] * √[2 - √2] / √2]*S =[√ [2√2 - √2 - 2 +√2] / √2]*S = [ √[2√2 -2]/ √2]*S= [√[(2)(√2 -1 )] / √2] *S = [√2 - 1]*S
So.... AB + AY = S + [√2 - 1]S = S [ 1 + [√2 - 1] ] S = √2S
Her's an (aproximate) picture......
Call the side of the square S.....
Using the Law of Cosines, we have
AX^2 = 2S^2 -2S^2cos(45) = 2S^2 - 2S^2(1/√2) = S^2 [ 2 - √2]
So........AX = S*√[ 2 - √2]
And using some basic geometry <YXD = 90 and <XDA = 45......so <XYD = 45....so <AYX = [180- <XYD]= 135
And since AB = BX and <ABD = 45, then <AXB = (180 - 45]/2 = 67.5
Then <AXY = [180 - 90 - 67.5] = 22.5
And using the Law of Sines, again, we have
AY/sin(22.5) = AX/sin(135)
AY = sin(22.5)/sin(135)* S*√[ 2 - √2] = [√[1-1/√2] / √2] * √2* √[2 - √2]S = [√[1-1/√2]*√[2 - √2]*S = [ √[√2 -1] * √[2 - √2] / √2]*S =[√ [2√2 - √2 - 2 +√2] / √2]*S = [ √[2√2 -2]/ √2]*S= [√[(2)(√2 -1 )] / √2] *S = [√2 - 1]*S
So.... AB + AY = S + [√2 - 1]S = S [ 1 + [√2 - 1] ] S = √2S
Her's an (aproximate) picture......