Question

[rororonoazoro]

Cho tam giác ABC có góc A=50^o, trực tâm H. Gọi D là điểm đối xứng với H qua BC.

a, CHứng minh tam giác BHC= tam giác BDC 

b, Tính góc BDC

Answer 1

[Nyatmax]

a.Goi BE,CG,AK la duong cao cua tam giac ABC\(\left(E\in AC,K\in BC,G\in AB\right)\)

Xet 2 tam giac vuong HKC va DKC ta co:

HK=DK(H doi xung voi D qua BC)

KC la canh chung

Do do:\(\Delta HKC=\Delta DKC\left(c-g-c\right)\)

Xet 2 tam giac vuong HKB va DKB ta co:

HK=DK(H doi xung voi D qua BC)

BK la canh chung

Do do:\(\Delta HKB=\Delta DKB\left(c-g-c\right)\)

Suy ra:\(\Delta HKB+\Delta HKC=\Delta KDB+\Delta KDC\left(1\right)\)

Ma:

\(\Delta BHC=\Delta HKB+\Delta HKC\left(2\right)\)

\(\Delta BDC=\Delta KDB+\Delta KDC\left(3\right)\)

Tu (1),(2) va (3) suy ra:\(\Delta BHC=\Delta BDC\)

b.Xet tam giac AGC ta co:\(\widehat{ACG}=40^0\)

Suy ra:\(\widehat{EHC}=50^0\)

Ma:\(\widehat{GHC}=\widehat{GHE}+\widehat{EHC}=180^0\)

\(\Rightarrow\widehat{GHE}=180^0-\widehat{EHC}=180^0-50^0=130^0\)

Hay \(\widehat{BHC}=130^0\)(\(\widehat{GHE}\)doi dinh \(\widehat{BHC}\))

Theo cau a ta co:\(\Delta BHC=\Delta BDC\)

Suy ra: \(\widehat{BHC}=\widehat{BDC}\)(2 goc tuong ung)

\(\Rightarrow\widehat{BDC}=130^0\)