Question

giúp với

Answer 1

Ta có hình vẽ sau:

Xét \(\Delta ABDvà\Delta KBD\) có:

BD: chung

\(\widehat{DBA}=\widehat{DBK}\left(gt\right)\)

AB = KB (gt)

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

\(\Rightarrow\widehat{BAD}=\widehat{BKD}=90^o\) (2 góc t/ứng)

Vậy......................

Answer 2

Xét \(\Delta ABD\) và \(\Delta KBD\) có:

\(AB=KB\left(gt\right)\)

\(\widehat{ABD}=\widehat{KBD}\) (tia pg)

BD chung

\(\Rightarrow\Delta ABD=\Delta KBD\left(c.g.c\right)\)

\(\Rightarrow\widehat{BAD}=\widehat{BKD}=90^o\)

Vậy \(\widehat{BKD}=90^o.\)

Answer 3

\(^{90^0}\)