問一題數學題~

According to the given figure,
∠BAD = ∠CBD and ∠ABD = ∠BCD

(a) ∠ADB = 180° -∠BAD -∠ABD

=180° - ∠CBD -∠BCD

=∠CDB

(b) draw a horizontal line DE so that

∠ADC =∠ADE+∠CDE

=(∠BAD+∠ABD) +(∠CBD+∠BCD)

= 2∠ABD+2∠CBD

= 2∠ABC

圖片參考：http://i181.photobucket.com/albums/x133/kissme1101/7.jpg

∠BAD = ∠CBD
∠ABD = ∠BCD
(a)
∠ADB
=180-∠BAD -∠ABD
=180- ∠CBD -∠BCD
=∠CDB
(b)
作一條水平線ＤＥ
∠ADC
=∠ADE+∠CDE
=(∠BAD+∠ABD) +(∠CBD+∠BCD)
= 2∠ABD+2∠CBD
=2∠ABC