Let M be the intersection of the diagonals AC and BD.
Let E and F be the feet of the perpendiculars from B
to AC and CD respectively.
ΔBFC is a 45-45 right triangle ⇒ |CF| = |BF|.
ΔBFD is a 60-30 right triangle ⇒ |BF| = |BM| = |MD| = |MF|.
|MD| = |MF| ⇒ ∠MFD = ∠MDF = 30°.
|CF| = |MF| ⇒ ∠FCM = ∠FMC = ½∠MFD = 15°.
∠FCM = 15° ⇒ ∠FCB = 30°.
ΔBEC is a 60-30 right triangle ⇒
|BE| = ½|BC| = ½|AD| = |AD|sin(30°). (1)
ΔBFC is a 45-45 right triangle ⇒ |BC| = |BF|√2.
ΔBFD is a 60-30 right triangle ⇒ |FD| = |BF|√3.
Therefore,
|AB| = |CD| = |CF| + |FD|
= |BF| + |BF|√3
= (1 + √3)|BF|
= (1 + √3)|BC|/√2 (2)
Combining (1) and (2) gives
|BE| = |AB|(√6 - √2)/4 = |AB|sin(15°).
QED
Note: See Harry's post for a trig. solution.
|
