Extend altitude BD to point F such that D is the midpoint of EF. Join
points C and F. Clearly, ΔCDF ≅ ΔCDE. Therefore, |CE| = |CF|
and ∠DCE = ∠DCF. Thus, ∠ECB = ∠ECF.
|BD| |AD|
(1) ------ = ------ [ ΔADB ~ ΔBDC ]
|BC| |AB|
|AD| |DE|
(2) ------ = ------ [ AE bisects ∠BAD ]
|AB| |BE|
|BD| |DE|
(3) ------ = ------ [ (1) & (2) ]
|BC| |BE|
2|BD| |FE|
(4) ------- = ------ [ (3) & |FE| = 2|DE| ]
|BC| |BE|
|FE| |CF|
(5) ------ = ------ [ CE bisects ∠BCF ]
|BE| |CB|
2|BD| |CF|
(6) ------- = ------ [ (4) & (5) ]
|BC| |BC|
|CE| = 2|BD| [(6) & |CE| = |CF| ]
QED
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