peterhu18
幼苗
共回答了13个问题采纳率:92.3% 举报
证明:1、
∵AB=AC
∴∠ABC=∠ACB
∵∠BAC=45
∴∠ABC=∠ACB=(180-45)/2=67.5
∵△A1B1C≌△ABC、
∴∠B1A1C=∠BAC=45,∠A1B1C=∠ABC=67.5,BC=B1C
∴∠BB1C=∠ABC=67.5
∴∠AB1A1=180-∠BB1C-∠A1B1C=180-67.5-67.5=45
∴∠AB1A1=∠B1A1C
∴AB//A1C
2、
△A1AB与△CBA1全等
证明:
∵AB=AC,AC=A1C
∴AB=A1C
∵AB//A1C
∴平行四边形ABCA1
∴AA1=BC
∵A1B=BA1
∴△A1AB与△CBA1全等
∵△A2B2C≌△ABC、
∴∠A2B2C=∠ABC=67.5
∴∠AB2A2=180-∠A2B2C=180-67.5=112.5
1年前
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peterhu18
△A1AB与△CBA1全等 证明: ∵AB//A1C ∴∠BA1C=∠ABA1 ∵AB=AC,AC=A1C ∴AB=A1C ∵A1B=BA1 ∴△A1AB与△CBA1全等(边角边相等)