Chứng minh BD vuông góc CK biết tam giác ABC vuông tại A có B=50 độ

\(a)Xet tam giac ABC, ta co:\)

\(\Rightarrow A+B+C=180^o\)

\(\Rightarrow90^o+50^o+C=180^o\)

\(\Rightarrow C=180^o-140^o\)

\(\Rightarrow C=40^o\)

\(b)XettamgiacABDvatamgiacEBD,taco:\)

\(\left\{{}\begin{matrix}AB=EB\left(gt\right)\\B_1=B_2\left(gt\right)\\BDchung\end{matrix}\right.\)

\(\Rightarrow\Delta ABD=\Delta EBD\left(c-g-c\right)\)

\(c)Taco: \bigtriangleup ABD = \bigtriangleup EBD\)

\(\Rightarrow\widehat{A}=\widehat{E}=90^o(2goctuongung)\)

\(\Rightarrow DE\perp BC\)

\(d)Taco:\)

\(A_1=E_1=90^o\Rightarrow A_2=E_2=90^o\)

\(Xet \bigtriangleup AKD va \bigtriangleup ECD, ta co:\)

\(\left\{{}\begin{matrix}A_2=E_2=90^o\\AD=ED\left(\Delta ABD=\Delta EBD\right)\\D_1=D_2\left(doidinh\right)\end{matrix}\right.\)

\(\Rightarrow\Delta ADK=\Delta EDC\left(g-c-g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}DK=DC\left(2canhtuongung\right)\\AK=EC\left(2canhtuongung\right)\end{matrix}\right.\)

\(e)Taco:\)

\(\left\{{}\begin{matrix}BA+AK=BK\\BE+EC=BC\\BA=BE\left(gt\right)\\AK=EC\left(\Delta ADK=\Delta EDC\right)\end{matrix}\right.\)

\(\Rightarrow BK=BC\)

\(Xet \bigtriangleup BKP va \bigtriangleup BCP,taco:\)

\(\left\{{}\begin{matrix}BK=BC\left(cmt\right)\\B_1=B_2\left(gt\right)\\BPchung\end{matrix}\right.\)

\(\Rightarrow\Delta BKP=\Delta BCP\left(c-g-c\right)\)

\(Xet\Delta DKPva\Delta DCP,taco:\)

\(\left\{{}\begin{matrix}KD=DC\left(\Delta ADK=\Delta EDC\right)\\DPchung\\KP=PC\left(\Delta BKP=\Delta BCP\right)\end{matrix}\right.\)

\(\Rightarrow\Delta DKP=\Delta DCP\left(c-c-c\right)\)

\(\Rightarrow P_1=P_2\left(2goctuongung\right)\)

\(Taco: P_1 + P_2 = 180^o (2gockebu)\)

\(Ma P_1 = P_2 (chungminhtren)\)

\(\Rightarrow P_1=P_2=\dfrac{180^o}{2}=90^o\)

\(\Rightarrow BP\bot CK\)