（2013•闸北区一模）如图，在△ABC中，AB=AC，∠A=36°，BD平分∠ABC交AC于点D，DE平分∠BDC交BC于点E，则ECAD=3−523−52．

Question

（2013•闸北区一模）如图，在△ABC中，AB=AC，∠A=36°，BD平分∠ABC交AC于点D，DE平分∠BDC交BC于点E，则(EC)/(AD)=

(3−√(5))/(2)

(3−√(5))/(2)

．(EC)/(AD)=

(3−√(5))/(2)

(3−√(5))/(2)

．(EC)/(AD)ECECADAD

(3−√(5))/(2)

(3−√(5))/(2)(3−√(5))/(2)3−√(5)3−√(5)√(5)

5

522

(3−√(5))/(2)

(3−√(5))/(2)(3−√(5))/(2)3−√(5)3−√(5)√(5)

5

522

Answer 1

∵在△ABC中，AB=AC，∠A=36°，∴∠ABC=∠C=72°，∵BD平分∠ABC交AC于点D，∴∠ABD=∠CBD=36°，∴∠ABD=∠A，∴AD=BD，∴∠BDC=180°-∠CBD-∠C=72°，∴∠BDC=∠C，∴BD=BC，∴AD=BD=BC，∵DE平分∠BDC交BC于点E...