Hình:
Giải:
Xét tam giác ABC, có:
\(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\) (Tổng ba góc tam giác)
\(\Leftrightarrow\widehat{ABC}+90^0+\widehat{BAC}=180^0\)
\(\Leftrightarrow\widehat{ABC}+\widehat{BAC}=90^0\)
\(\Leftrightarrow\widehat{ABC}=90^0-\widehat{BAC}\) (1)
Lại xét tam giác ABD, có:
\(\widehat{ABD}+\widehat{ADB}+\widehat{BAD}=180^0\) (Tổng ba góc tam giác)
\(\Leftrightarrow90^0+\widehat{ADB}+\widehat{BAC}=180^0\)
\(\Leftrightarrow\widehat{ADB}+\widehat{BAC}=90^0\)
\(\Leftrightarrow\widehat{ABD}=90^0-\widehat{BAC}\) (2)
Từ (1) và (2) \(\Rightarrow\widehat{ABC}=\widehat{ADB}\) (Bắc cầu)
Xét tam giác DBC, có:
\(\widehat{DBC}+\widehat{DCB}+\widehat{BDC}=180^0\) (Tổng ba góc tam giác)
\(\Leftrightarrow\widehat{DBC}+90^0+\widehat{BDC}=180^0\)
\(\Leftrightarrow\widehat{DBC}+\widehat{BDC}=90^0\)
\(\Leftrightarrow\widehat{DBC}=90^0-\widehat{BDC}\) (3)
Lại xét tam giác DAB, có:
\(\widehat{ABD}+\widehat{ADB}+\widehat{DAB}=180^0\) (Tổng ba góc tam giác)
\(\Leftrightarrow90^0+\widehat{BDC}+\widehat{DAB}=180^0\)
\(\Leftrightarrow\widehat{BDC}+\widehat{DAB}=90^0\)
\(\Leftrightarrow\widehat{DAB}=90^0-\widehat{BDC}\) (4)
Từ (3) và (4) \(\Rightarrow\widehat{DBC}=\widehat{DAB}\) (Bắc cầu)
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