Question

初一数学，帮帮忙！已知.AB‖CD,AM,CM分别平分∠BAP,∠DCP.试探索∠M 与∠P 的数量关系.

Answer 1

1.
∠M=180-∠MAC-∠MCA=180-(∠BAC-∠BAM)-(∠DCA-∠DCM)
=180-(∠BAC+∠DCA)+∠BAM+∠DCM=∠BAM+∠DCM=∠MAP+∠MCP

∠P=180-∠PAC-∠PCA=180-(∠BAC-∠BAP)-(∠DCA-∠DCP)
=180-(∠BAC+∠DCA)+∠BAP+∠DCP=∠BAP+∠DCP=2(∠MAP+∠MCP)=2∠M

2.
∠M=180-∠MAC-∠MCA=180-(∠BAC-∠BAM)-(∠DCA-∠DCM)
=180-(∠BAC+∠DCA)+∠BAM+∠DCM=∠BAM+∠DCM=∠MAP+∠MCP
∠P=180-∠PAC-∠PCA=180-(∠BAP-∠BAC)-(∠DCP-∠DCA)
=180+(∠BAC+∠DCA)-(∠BAP+∠DCP)=360-∠BAP-∠DCP=360-2(∠MAP+∠MCP)=360-2∠M
3.
∠M=180-∠MAC-∠MCA=180-(∠BAC+∠BAM)-(∠DCA-∠DCM)
=180-(∠BAC+∠DCA)-∠BAM+∠DCM=-∠BAM+∠DCM=-∠MAP+∠MCP
∠P=180-∠PAC-∠PCA=180-(∠BAC+∠BAP)-(∠DCA-∠DCP)
=180-(∠BAC+∠DCA)-∠BAP+∠DCP=-∠BAP+∠DCP=2(-∠MAP+∠MCP)=2∠M

Answer 2

第一个图 ∠2m＝∠p