Dao Trong Luan Coordinator
25/12/2017 at 11:41
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a) Apply properties of certain segments , we have :
AD // BM
AB // MD
=> AD = BM
Similar , we have EA = MC
=> AD + EA = BM + MC
=> DE = BC (1)
We can see : \(\angle EDM=\angle ABC\) (Because ED // BM , AD // MB)
\(\angle DEM=\angle ACB\) (Because ED // BM , EM // AC)
=> \(\Delta MED=\Delta ACB\left(e-a-e\right)\)
Selected by MathYouLike -
Faded 22/01/2018 at 12:42a) Apply properties of certain segments , we have :
AD // BM
AB // MD
=> AD = BM
Similar , we have EA = MC
=> AD + EA = BM + MC
=> DE = BC (1)
We can see : ∠EDM=∠ABC
(Because ED // BM , AD // MB)
∠DEM=∠ACB
(Because ED // BM , EM // AC)
=> ΔMED=ΔACB(e−a−e)