標題:
math problems HELP~!!
發問:
In Figure, ABCD is a parallelogram. EBDF is a straight line and EB = DF. (a) Prove that ∠ABE = ∠CDF. (b) Prove that EA // CF. Figure http://server6.pictiger.com/img/159765/picture-hosting/34.jpg
最佳解答:
(a) To prove ∠ABE = ∠CDF: ∠ABD=∠CDB (alt. ∠s, AB//CD) ∠ABE+∠ABD=180 (adj. ∠s on st. line) ∠ABE=180-∠ABD ∠CDF+∠CDB=180 (adj. ∠s on st. line) ∠CDF=180-∠CDB ∠CDF=180-∠ABD=∠ABE (b) To prove EA//CF: Comparing triangle EAB and triangle FCD EB=DF (given) ∠ABE = ∠CDF (proved in (a)) AB=CD (given) so triangle EAB (~ =)(congus.) triangle FCD ∠BEA = ∠DFC (corr. ∠s, ~= trianlges) so EA // CF (alt. ∠s equal)
其他解答:
a) draw CD upwards longer and let G be on the produced line ∠ABE = ∠BDG ∠BDG = ∠CDF so, ∠ABE = ∠CDF b) EB = DF (given) AB = CD and ∠ABE = ∠CDF triangle ABE and triangle CDF are congruent ∠AEB = ∠CFD so, EA // CF
math problems HELP~!!
發問:
In Figure, ABCD is a parallelogram. EBDF is a straight line and EB = DF. (a) Prove that ∠ABE = ∠CDF. (b) Prove that EA // CF. Figure http://server6.pictiger.com/img/159765/picture-hosting/34.jpg
最佳解答:
(a) To prove ∠ABE = ∠CDF: ∠ABD=∠CDB (alt. ∠s, AB//CD) ∠ABE+∠ABD=180 (adj. ∠s on st. line) ∠ABE=180-∠ABD ∠CDF+∠CDB=180 (adj. ∠s on st. line) ∠CDF=180-∠CDB ∠CDF=180-∠ABD=∠ABE (b) To prove EA//CF: Comparing triangle EAB and triangle FCD EB=DF (given) ∠ABE = ∠CDF (proved in (a)) AB=CD (given) so triangle EAB (~ =)(congus.) triangle FCD ∠BEA = ∠DFC (corr. ∠s, ~= trianlges) so EA // CF (alt. ∠s equal)
其他解答:
a) draw CD upwards longer and let G be on the produced line ∠ABE = ∠BDG ∠BDG = ∠CDF so, ∠ABE = ∠CDF b) EB = DF (given) AB = CD and ∠ABE = ∠CDF triangle ABE and triangle CDF are congruent ∠AEB = ∠CFD so, EA // CF
此文章來自奇摩知識+如有不便請留言告知
文章標籤
全站熱搜
留言列表
