Question

giúp ạ!!!!!

Answer 1

a: Qua C, kẻ CH⊥AB tại H

Ta có: Ax⊥ AB

CH⊥AB

By⊥BA

Do đó: Ax//CH//By

Ta có: Ax//CH

=>\(\hat{xAC}=\hat{ACH}\) (hai góc so le trong)

ta có: CH//By

=>\(\hat{HCB}=\hat{CBy}\) (hai góc so le trong)

\(\hat{ACB}=\hat{ACH}+\hat{BCH}\)

\(=\hat{xAB}+\hat{yBC}\)

\(\hat{ACB}+\hat{CAB}+\hat{CBA}\)

\(=\hat{xAB}+\hat{CAB}+\hat{yBC}+\hat{CBA}\)

\(=90^0+90^0=180^0\)

b: \(\hat{DAB}+\hat{DBA}=\hat{DAC}+\hat{DBC}+\hat{CAB}+\hat{CBA}\)

\(=\frac12\left(\hat{xAC}+\hat{yBC}\right)+180^0-\hat{ACB}\)

\(=\frac12\cdot\hat{ACB}+180^0-\hat{ACB}=180^0-\frac12\cdot\hat{ACB}\)

=>\(\hat{ADB}=180^0-\left(180^0-\frac12\cdot\hat{ACB}\right)=\frac12\cdot\hat{ACB}\)

=>\(\hat{ACB}=2\cdot\hat{ADB}\)