2个问的啊thx10points

1)ㄥBEC = ㄥAED = 90° (given)
BE=DE (given)
BC=DA (given)△BEC≌△DAE (R.H.S.)
2)Consider △DFC and △BEC :
ㄥDCF = ㄥBCF (common)
△BEC≌△DAE (proved) so ㄥCBE = ㄥCDF (corr. ∠s, ~ Δs)△DFC ~ △BEC (A.A.)So ㄥDFC = ㄥBEC = 90°ㄥBFA + ㄥDFC = 180° (adj. ∠s on st. line)
ㄥBFA + 90° = 180°
ㄥBFA = 90°


2011-01-30 13:59:12 補充：
Typing error :

For Q2 ,
ㄥDCF = ㄥBCF (common)
should be
ㄥDCF = ㄥBCE (common)