Question

如图,在△ABC中,AB=AC,∠ABC=72°(1)用直线和圆规作∠ABC的平分线BD交AC于点D

如图,在△ABC中,AB=AC,∠ABC=72°(1)用直线和圆规作∠ABC的平分线BD交AC于点D(2)在(1)中作出∠ABC的平分线BD后,求∠BDC的度数

Answer 1

1)以B为圆心，Ba为半径作弧，交AB于a，交BC于b，分别a、b为圆心ao为半径作弧，两弧交于o点，连接BO并延长交AC于D.

2)∵BD平分∠ABC

∴∠CBD=1/2∠ABC=36°

∵AB=AC

∴∠C=∠ABC=72°

∴∠BDC=180-(∠CBD+∠C)

              =180-(36+72)

              =72°