在三角形ABC中，AB=AC,CD⊥AB交AB于D,∠ABC的平分线BE交CD于E,求∠BEC的度数
\u5982\u56fe\uff0e\u25b3ABC\u4e2d\uff0cAB=AC\uff0cCD\u22a5AB\u4ea4AB\u4e8eD\uff0c\u2220ABC\u7684\u5e73\u5206\u7ebfBE\u4ea4CD\u4e0eE\uff0c\u5219\u2220BEC\u7684\u5927\u5c0f\u662f\uff08\u3000\u3000\uff09 A\uff0e135\u00b0-\u2235AB=AC\uff0c\u2234\u2220ABC=\u2220ACB= 1 2 \uff08180\u00b0-\u2220A\uff09=90\u00b0- 1 2 \u2220A\uff0c\u53c8\u2235BE\u662f\u2220ABC\u7684\u5e73\u5206\u7ebf\uff0c\u2234\u2220DBE= 1 2 \u2220ABC=45\u00b0- 1 4 \u2220A\uff0e\u2235\u2220BEC\u662f\u25b3BED\u7684\u5916\u89d2\uff0cCD\u22a5AB\uff0c\u2234\u2220BEC=\u2220BDE+\u2220DBE=90\u00b0+45\u00b0- 1 4 \u2220A=135\u00b0- 1 4 \u2220A\uff0e\u6545\u9009A\uff0e
\u8bbeCE\u4e0eBD\u7684\u4ea4\u70b9\u4e3aO
\u8fde\u63a5ED\uff0c\u5219\u7531\u89d2\u5e73\u5206\u7ebf\u7684\u6027\u8d28\uff1a
\uff08\u6211\u5728\u63d0\u95ee\uff1ahttp://zhidao.baidu.com/question/367380732249565244.html\uff0c\u4e2d\u5df2\u7ecf\u8bf4\u660e\uff09
BE/BC=EO/EC
CD/CB=DO/OB
\u2235 BE=CD
\u2234 EO/OC=DO/OB
\u2234 ED//BC
\u2234 AD/DC=AE/EB\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2460
\u7531\u89d2\u5e73\u5206\u7ebf\u7684\u6027\u8d28\uff1a
BA/BC=AD/DC
AC/BC=AE/EB
\u4e14\u7531\u2460
AD/DC=AE/EB
\u2234 BA/BC=AC/BC
\u2234 AB=AC
\u3010\u8bf4\u660e\u3011\u672c\u9898\u4e5f\u53ef\u76f4\u63a5\u8bc1\u660e\uff1aAE=AD
\u8bbeCE\u4e0eBD\u7684\u4ea4\u70b9\u4e3aO
\u8fde\u63a5ED\uff0c\u5219\u7531\u89d2\u5e73\u5206\u7ebf\u7684\u6027\u8d28\uff1a
BE/BC=EO/EC
CD/CB=DO/OB
\u2235 BE=CD
\u2234 EO/OC=DO/OB
\u2234 ED//BC
\u2234 AD/DC=AE/EB\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2460
\u2235 BE=CD
\u2234 AD=AE
\u2234 AB=AC
∵AB=AC
∴∠C=∠ABC=(180°-∠A)/2
∵BE平方∠ABC
∴∠DBE=1/2×∠ABC=1/4×﹙180°-∠A)
∵CD⊥AB
∴∠BDC=90°
∴∠BEC=90°+1/4×﹙180°-∠A)
=135°-1/4∠A
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绛旓細鍙В3闂鍏朵粬鏄撶煡閬擄紱璁捐A宸茬煡锛涘垯瑙払=(180-A)/2鍒欒NMB=90-瑙払=90-锛180-A锛/2=A/2
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