Let ∠ABC of ΔABC be a right angle and the bisector of angle ∠BAC
intersect the altitude BD at point E. Prove the following:

If m(∠BCE) = 2*m(∠ACE), then |CE| = 2*|BD|.
  

Use c to denote /DCE, so that /BCE = 2c, /ABE = 3c, /BEC = 90^o + c.

|BE|/|DE| =  |AB|/|AD| = 1/sin(3c)         since AE bisects angle A.

Also, |BE|/|DE|  = (|BE|/|CE|)*(|CE|/|DE|)

                        = (|BE|/|CE|)/sin(c)

                        = sin(2c)/(cos(3c)*sin(c))           using sine rule in BCE.

Equating these results:   sin(2c)/(cos(3c)*sin(c)) = 1/sin(3c),   giving

tan(3c) = sin(c)/sin(2c) = sin(c)/(2sin(c)cos(c)) = 1/(2cos(c)).       (1)

Also, |BD|/|CE|  = (|BD|/|BC|)*(|BC|/|CE|)

                        =  sin(3c)*cos(c)/cos(3c)            using sine rule in BCE.

                        = cos(c)tan(3c)

                        = cos(c)/(2cos(c))                      using (1)

                        = 1/2

Thus |CE| = 2*|BD|

I wonder if there’s a simpler way without the trigonometry?

Posted by Harry on 2014-04-02 18:11:53