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问题: 关于相似三角形的题目||||||

在△ABC中,∠ACB-90°,CH⊥AB于H,△ACD和△BCE均为等边三角形。
(1)求证:△DAH∽△ECH;
(2)若AH:HB=1:4,求S△DAH:S△ECH

解答:

CH⊥AB于H
==>△ACH∽△ABC
CH:AH =BC:AC.....(1)
△ACD和△BCE均为等边三角形
==>BC:AC=AD:CE .....(2)
(1),(2)==>CH:AH =AD:CE ......(3)
显然:∠CAH=∠HCB...(4)
又∠DAC=∠BCE=60度....(5)
(4)+(5)==>∠DAH=∠HCE....(6)
(3),(6)==>△DAH∽△ECH

2)
AH:HB=1:4==>AH:AB=1:5
△ACH∽△ABC
AH:AC =AC:AB
AC^2 =AH*AB =5AH^2 .......(7)
AB^2=(6AH)^2=36AH^2
BC^2 =AB^2-AC^2 =31AH^2 ....(8)
(7):(8)==>5:31
S△DAH:S△ECH
=(AD:CE)^2 =AD^2:CE^2=AC^2:AB^2 =5:31