2024-12-25 15:27| 查看: 1
≌;AB和BE, BC和EF,AC和DF
(2)
30°;≌;∠E和∠C, ∠D和∠B, ∠DAE,∠BAC
3.
解:
∵△ABC≌△BAD,BC与AD足对应边
∴AB=BA,AD=BC,BD=AC,
∠D=∠C,∠DAB=∠CBA, ABD=∠BAC
4.
解:∵△DEF≌△PQK
∴PQ=DE=5cm,QK=EF=8cm,PK=DF=7cm
∴QK-FK=EF-FK,即QF=EK=3cm
∴FK=EF-EK=5cm,EQ=EF+FQ=11cm