角ACB等于3角B,角1等于角2,CD垂直AD于D,求证：AB—AC=2CD

Question

角ACB等于3角B,角1等于角2,CD垂直AD于D,求证：AB—AC=2CD

Answer 1

求：AB-AC=2CD

延长CD至点E E在AB上

 在RT△ADE与RT△ADC中：∠1=∠2    AD=AD    ∠ADE=∠ADC=90°

 ∴△ADE≌△ADC(ASA) ∴∠AED=∠ACD   且AE=AC 

 ∴∠ACB+∠ABC=∠ACE+∠AEC 

 ∴∠AEC=∠ABC+∠ECB 

又∵∠ACB=3∠B 

   ∴∠B=∠ECB ∴EB=EC 

   ∵EC=ED+CD=2CD 

   ∴AB=AE+EB=AC+EC=AC+2CD

即  AB-AC=2CD