a) Xet △ABE va △HBE co:
\(\left\{{}\begin{matrix}AB=BH\\B_1=B_2\\chugBE\end{matrix}\right.\)
=> △ABE=△HBE(c.g.c)
=> goc BAE= goc BHE = 90 do
=> EH⊥BC
b) CM duoc △ABD=△BHD(c.g.c)
=> goc BDA= goc BDH (GTU) va goc BDA+goc BDH=180
=> goc BDA=goc BDH=90
=> BD⊥AH (1)
Vi △ABD=△BHD => AD=DH va D∈AH => D là trung điểm của AH (2)
Tu (1)(2) => dpcm
c) Xet △AEK va △HEC co:
\(\left\{{}\begin{matrix}E_1=E_2\left(d^2\right)\\AE=EH\\gocKAE=gocCHE=90\end{matrix}\right.\)
=> △AEK=△HEC(g.c.g)
=> EK=EC(CTU)
d) Dung duong trung binh
e) Ta co : \(\left\{{}\begin{matrix}E_2=E_1\\BEA=MEC\left(d^2\right)\\BEH=KEM\left(d^2\right)\end{matrix}\right.\)
=> E2+BEA+BEH=E1+MEC+KEM (1)
Vi E∈AC nen E2+BEA+BEH=180 (2)
Tu(1)(2) => E1+MEC+KEM=180 => B,E,M thang hang
