Question

如图 圆o是△abc的外接圆 bc为直径,ad平分角bac,交圆o于d,点m是△abc的内心

Answer 1

①BC=√2DM.

∵∠BMD=∠BAM+∠ABM

=1/2(∠BAC+∠ABC)

∠MBD=∠MBC+∠DBC

=1/2∠ABC+∠DAC

=1/2(∠ABC+∠BAC),

∴∠MBD=∠BMD,

∴BD=MD,

连接CD,

BC中直径,

∴∠BDC=90°,

又BD=CD