In the triangle ABC, CD is high, e is the midpoint of CD, the extension line of AE intersects CB at point F, AE = CE and AF is vertical to CB. If EF = 1, then AE
CD⊥AB,AF⊥BC
∠CDA=∠CFA=90
∵AE=CE
∴∠FAC=∠DCA
∠ECF=∠EAD
△ECF∽△EAD
AE:EF=DE:CE
E is the midpoint of CD, so de = CE
EF=1
So AE: 1 = 1
AE=1
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