Question

急如图,AE,AD是直线且AB＝BC＝CD＝DE＝EF＝FG＝GA,若∠DAE＝a,求a的值 急,

Answer 1

∠ACB＝∠DAE＝a,

由外角知∠CBD＝∠CDB＝∠ACB+∠DAE＝2a;

∠AED＝∠ADE＝∠DEC,

∠EDC＝∠DAE＝a;所以∠AED＝∠ADE＝∠CDE+∠CDA＝3a

所以∠DAE+∠ADE+∠AED＝a+3a+3a=7a=180度

所以a=180/7度