a) Xét $∆ABK$ và $∆ACK$ có:
$AB = AC\quad (gt)$
$KB = KC =\dfrac12BC\quad (gt)$
$AK:$ cạnh chung
Do đó $∆ABK=∆ACK\, (c.c.c)$
$\Rightarrow \widehat{AKB}=\widehat{AKC}$ (hai góc tương ứng)
mà $\widehat{AKB}+\widehat{AKC}=180^\circ$ (hai góc kề bù)
nên $\widehat{AKB}=\widehat{AKC} = 90^\circ$
$\Rightarrow AK\perp BC$
b) Ta có: $EC\perp BC\quad (gt)$
$AK\perp BC$ (câu a)
nên $EC//AK\quad (\perp BC)$
c) Ta có:
$∆ABK=∆ACK$ (câu a)
$\Rightarrow \widehat{KAB}=\widehat{KAC}$ (hai góc tương ứng)
mà $\widehat{KAB}+\widehat{KAC}=\widehat{BAC}=30^\circ$
nên $\widehat{KAB}=\widehat{KAC} = 15^\circ$
Ta có: $EC//AK$ (câu b)
$\Rightarrow \widehat{KAB}=\widehat{BEC}$ (hai góc đồng vị)
Do đó: $\widehat{BEC}=15^\circ$
