Đáp án:

a)ΔABD;AB=BD

⇒ΔABD⇒ΔABD cân tại BB

⇒ˆBAD=ˆBDAb)AB⊥ACDK⊥AC⇒AB//DK⇒ˆBAD=ˆD1⇒BAD^=BDA^b)AB⊥ACDK⊥AC⇒AB//DK⇒BAD^=D1^

Mà ˆBAD=ˆBDABAD^=BDA^

⇒ˆD1=ˆBDA⇒D1^=BDA^

 ΔAHDΔAHD và ΔAKDΔAKD

AD:AD: chung

ˆAHD=ˆAKD=90oˆD1=ˆBDA⇒ΔAHD=ΔAKD⇒A1=ˆA2AHD^=AKD^=90oD1^=BDA^⇒ΔAHD=ΔAKD⇒A1=A2^

⇒AD⇒AD là phân giác ˆHACHAC^

c)⇒ΔAHD=ΔAKD⇒AH=AKd)AB+AC=AB+AK+KC<BD+AH+DC=AH+BC

$a)\Delta ABD; AB=BD$

$\Rightarrow \Delta ABD$ cân tại $B$

$\Rightarrow \widehat{BAD}=\widehat{BDA}\\ b)AB \perp AC\\ DK \perp AC\\ \Rightarrow AB//DK\\ \Rightarrow \widehat{BAD}=\widehat{D_1}$

Mà $\widehat{BAD}=\widehat{BDA}$

$\Rightarrow \widehat{D_1}=\widehat{BDA}$

Xét $\Delta AHD$ và $\Delta AKD$

$AD:$ chung

$\widehat{AHD}=\widehat{AKD}=90^o\\ \widehat{D_1}=\widehat{BDA}\\ \Rightarrow \Delta AHD = \Delta AKD\\ \Rightarrow {A_1}=\widehat{A_2}$

$\Rightarrow AD$ là phân giác $\widehat{HAC}$

$c)\Rightarrow \Delta AHD = \Delta AKD\\ \Rightarrow AH=AK\\ d)AB+AC=AB+AK+KC<BD+AH+DC=AH+BC$