ABCD is a square, AC & BD are its diagonals which intersect at O. Given angle BDL =angle LDC, and BL = 6cm. find the length of the line segment OK.
Call S the side of the square
Using the Law of Sines
LC/ sin (LDC) = S / sin (67.5)
(S - 6) /sin 22.5) = S / sin (67.5)
(S - 6) / S = sin (22.5) / sin (67.5)
(S - 6) /S = sin (22.5) / cos(22.5)
(S - 6) /S = tan (22.5)
(S - 6) / S = tan (45/2)
(S - 6) / S = ( 1 - cos(45) ) / sin (45)
(S - 6) / S = ( 1 - √2/2 ) / ( √2/2)
(S - 6) / S = [ 2 - √2] / √2
(S - 6) / S = √2 - 1
S - 6 = S [√2 - 1]
S - S [ √2 -1] = 6
S [ 2 - √2 ] = 6
S = 6 / [ 2 - √2]
S = 6 [ 2 +√2] / 2 = 3 [ 2 +√2] = 6 + 3√2
And applying the Law of Sines again
S / sin (DKC) = KC / sin(KDC)
S / sin (112.5) = KC / sin (22.5)
KC = S* sin ( 22.5) / sin (112.5)
KC = S * sin (22.5) / sin (67.5)
KC = S * sin (22.5) / cos(22.5)
KC = S * tan (22.5)
KC = S * [√2 - 1]
KC = 3 [ 2 + √2] [ √2 - 1]
KC = 3 [ 2√2 + 2 - 1 - √2]
KC = 3 √2
So the length of 1/2 of the diagonal = OC =
S√2 / 2 = ( [ 6 + 3√2] * √2 ) /2 = [ 6√2 + 6 ] / 2 =
3√2 + 3
So
OK = OC - KC = [ 3√2 + 3 ] - 3√2 = 3