a) Xét \(\Delta ADB\)và \(\Delta ADC\)có :

AD ( cạnh chung )

\(\widehat{A_1}=\widehat{A_2}\)( vì AD là tia phân giác )

AB = AC ( gt )

suy ra \(\Delta ADB\)= \(\Delta ADC\)( c.g.c )

b) \(\Rightarrow\widehat{ADB}=\widehat{ADC}\)( 2 góc tương ứng )                         ( theo câu a )

Mà \(\widehat{ADB}+\widehat{ADC}=180^o\)

\(\Rightarrow\widehat{ADB}=\widehat{ADC}=\frac{180^o}{2}=90^o\)

\(\Rightarrow AD\perp BC\)

c) vì \(\Delta ADB\)= \(\Delta ADC\)( theo câu a )

\(\Rightarrow BD=CD\)( 2 cạnh tương ứng )

\(\Rightarrow\widehat{ABD}=\widehat{ACD}\)( 2 góc tương ứng )

Mà \(\widehat{ABD}+\widehat{BDH}=90^o\); \(\widehat{ACD}+\widehat{CDK}=90^o\)

\(\Rightarrow\widehat{BDH}=\widehat{CDK}\)

Xét \(\Delta HBD\)và \(\Delta KCD\)có :

\(\widehat{BDH}=\widehat{CDK}\)( cmt )

BD = CD ( cmt )

\(\widehat{ABD}=\widehat{ACD}\)( cmt )

suy ra \(\Delta HBD\)= \(\Delta KCD\)( g.c.g )

\(\Rightarrow DH=DK\)( 2 cạnh tương ứng )

Giúp mình với nhé.thanhk you!!!

\(\frac{29}{110}nha.\)

Tham khảo here =))

https://olm.vn/hoi-dap/detail/67509118574.html

12 ×(-2/3)^2 +4/3 = 20/3 . 

=12*(\(\frac{-2}{3}\))^2+\(\frac{4}{3}\)

=12*\(\frac{4}{9}\)+\(\frac{4}{3}\)

=\(\frac{16}{3}\)+\(\frac{4}{3}\)

=\(\frac{20}{3}\)

\(\Delta\)DEF có góc E = góc F nên \(\Delta\)DEF cân tại D. Suy ra DE=DF