Question

如图,在三角形ABC中,AB=AC,D是三角形ABC外一点,且角ABD=角ACD=60度,求证；BD+DC=AB

Answer 1

延长BD到E,使DE=CD,连结AE

∵∠ABD=∠ACD

∴四边形ABCD四点共圆

则∠CAD=∠CBD ∠BAC=∠BDC ∠ACB=∠ADB

又∵AB=AC

∴∠ABC=∠ACB=∠ADB

∴∠ADE

=∠ABD+∠BAD

=∠ABD+∠CAD+∠BAC

=∠ABD+∠CBD+∠BDC

=∠ABC+∠BDC

=∠ADB+∠BDC

=∠ADC

在△ADE和△ADC中

AD=AD,∠ADE=∠ADC,DE=DC

∴△ADE≌△ADC

∠AED=∠ACD=60°

那么在△ABE中

∠ABE=∠AEB=60°

△ABE是等边三角形

∴AB=BE=BD+DE=BD+CD

=∠ADB+∠BDC

=∠ADC