Đáp án: LẦN NÀY KHO ĐÒI VẼ HÌNH NỮA AK
Ta có: ADAB=DCBCADAB=DCBC (t/c)⇒AD4=DC6=AD+DC4+6=510=12⇒AD4=DC6=AD+DC4+6=510=12
⇒AD=4.12=2,DC=6.12=3⇒AD=4.12=2,DC=6.12=3.
Suy ra
DIIB=DCCB=36=12⇒DIDB=13DIIB=DCCB=36=12⇒DIDB=13;
BEEA=BCAC=65⇒BEBA=611BEEA=BCAC=65⇒BEBA=611;
ADDC=23⇒ADAC=25ADDC=23⇒ADAC=25.
Suy ra SDIE=13SBDE,SDIE=13SBDE, SDBE=611SABD,SDBE=611SABD, SABD=25SABCSABD=25SABC
⇒SDIE=13.611.25=455SABC⇒SDIE=13.611.25=455SABC.
Vậy SDIESABC=455SDIESABC=455.
OK NHA ANH HAI