Đáp án:

$\widehat{BKC}={110}^o$

Giải thích các bước giải:

a) Ta có:

$K$ đối xứng với $H$ qua $BC$

$\Rightarrow BC$ là trung trực của $HK$

$\Rightarrow BH=BK;CH=CK$

Xét $\Delta BHC$ và $\Delta BKC$ có:

$BH=BK\left(cmt\right)$

$CH=CK\left(cmt\right)$

$BC:$ cạnh chung

Do đó $\Delta BHC=\Delta BKC(c.c.c)$

b) Ta có:

$\widehat{BHK}=\widehat{BAH}+\widehat{ABH}$ (góc ngoài của $\Delta ABH$)

$\widehat{CHK}=\widehat{CAH}+\widehat{ACH}$ (góc ngoài của $\Delta ACH$)

$\Rightarrow \widehat{BHC}=\widehat{BHK}+\widehat{CHK}$

$=\widehat{BAH}+\widehat{ABH}+\widehat{CAH}+\widehat{ACH}$

$=\widehat{BAC}+\widehat{ABH}+\widehat{ACH}$

Ta lại có:

$\widehat{BAC}+\widehat{ABH}={90}^o$ $(BH\perp AC)$

$\widehat{BAC}+\widehat{ACH}={90}^o$ $(CH\perp AB)$

$\Rightarrow 2\widehat{BAC}+\widehat{ABH}+\widehat{ACH}={180}^o$

$\Rightarrow \widehat{ABH}+\widehat{ACH}={180}^o-2\widehat{BAC}$

Do đó:

$\widehat{BHC}=\widehat{BAC}+{180}^o-2\widehat{BAC}={180}^o-\widehat{BAC}={180}^o-{70}^o={110}^o$

Mặt khác:

$\widehat{BHC}=\widehat{BKC}(\Delta BHC=\Delta BKC)$

$\Rightarrow \widehat{BKC}={110}^o$

Đáp án:

$\widehat{BKC}={110}^o$

Giải thích các bước giải:

a) Ta có:

$K$ đối xứng với $H$ qua $BC$

$\Rightarrow BC$ là trung trực của $HK$

$\Rightarrow BH=BK;CH=CK$

Xét $\Delta BHC$ và $\Delta BKC$ có:

$BH=BK\left(cmt\right)$

$CH=CK\left(cmt\right)$

$BC:$ cạnh chung

Do đó $\Delta BHC=\Delta BKC(c.c.c)$

b) Ta có:

$\widehat{BHK}=\widehat{BAH}+\widehat{ABH}$ (góc ngoài của $\Delta ABH$)

$\widehat{CHK}=\widehat{CAH}+\widehat{ACH}$ (góc ngoài của $\Delta ACH$)

$\Rightarrow \widehat{BHC}=\widehat{BHK}+\widehat{CHK}$

$=\widehat{BAH}+\widehat{ABH}+\widehat{CAH}+\widehat{ACH}$

$=\widehat{BAC}+\widehat{ABH}+\widehat{ACH}$

Ta lại có:

$\widehat{BAC}+\widehat{ABH}={90}^o$ $(BH\perp AC)$

$\widehat{BAC}+\widehat{ACH}={90}^o$ $(CH\perp AB)$

$\Rightarrow 2\widehat{BAC}+\widehat{ABH}+\widehat{ACH}={180}^o$

$\Rightarrow \widehat{ABH}+\widehat{ACH}={180}^o-2\widehat{BAC}$

Do đó:

$\widehat{BHC}=\widehat{BAC}+{180}^o-2\widehat{BAC}={180}^o-\widehat{BAC}={180}^o-{70}^o={110}^o$

Mặt khác:

$\widehat{BHC}=\widehat{BKC}(\Delta BHC=\Delta BKC)$

$\Rightarrow \widehat{BKC}={110}^o$

a.

Ta có:

\(\widehat{A}+\widehat{D}=180^0\Rightarrow\widehat{D}=180-\widehat{A}=180^0-120^0=60^0\\ \widehat{B}+\widehat{C}=180^0\Rightarrow\widehat{B}=180^0-\widehat{C}=180^0-80^0=100^0\)

b.

Độ dài đường trung bình MN của hình thang là:

\(MN=\dfrac{AB+CD}{2}=\dfrac{5+12}{2}=\dfrac{17}{2}cm\)

Đs....