Question

如图，已知：在直角△ABC中，∠C=90°，BD平分 ∠ABC交AC于D

1）若AD=BD，求证：∠BAC的度数（2）若AP平分∠ABC交BD于P，求∠BPA的度数

Answer 1

(1)∵AD=BD
∴∠ABD=∠BAD
∵BD平分∠ABC
∴2∠ABD=∠ABC
∵∠ABC+∠BAD=90º
∴3∠ABD=90º，则∠ABD=30º
即：∠BAC=30º

追答

(2)∵BD平分∠ABC
∴∠ABD=(1/2)∠ABC
同理：∠BAP=(1/2)∠BAC
∴∠ABD+∠BAP=(1/2)(∠ABC + ∠BAC)
=(1/2)×90º=45º
∴∠BPA=180º-(∠ABD+∠BAP)
=180º-45º=135º

Answer 2

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