Bài 3: Cho ΔABC cân có AB = AC = 5cm, BC = 8cm. Kẻ AH vuông góc BC (H thuộc BC) a. Chứng minh: HB = HC. b. Tính độ dài AH. c. Kẻ HD vuông góc với AB (D∈AB), kẻ HE vuông góc với AC (E∈AC). Chứng minh ΔHDE cân. d) So sánh HD và HC. Bài 4. Cho tam giác ABC có AB = AC = 5cm, BC = 6cm

thanhnghe75

2 năm trước

Bài 3: Cho ΔABC cân có AB = AC = 5cm, BC = 8cm. Kẻ AH vuông góc BC (H thuộc BC) a. Chứng minh: HB = HC. b. Tính độ dài AH. c. Kẻ HD vuông góc với AB (D∈AB), kẻ HE vuông góc với AC (E∈AC). Chứng minh ΔHDE cân. d) So sánh HD và HC. Bài 4. Cho tam giác ABC có AB = AC = 5cm, BC = 6cm. Đường trung tuyến AM xuất phát từ đỉnh A của tam giác ABC. a) Chứng minh ΔAMB = ΔAMC và AM là tia phân giác của góc A. b) Chứng minh AM c) Tính độ dài các đoạn thẳng BM và AM. d) Từ M vẽ ME AB (E thuộc AB) và MF AC (F thuộc AC). Tam giác MEF là tam giác gì? Vì sao?

Loga Sinh Học lớp 12

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duann22

Giải thích các bước giải:

Bài 3:

a.Xét $\Delta AHB, \Delta AHC$ có:

Chung $AH$

$\widehat{AHB}=\widehat{AHC}(=90^o)$
$AB=AC$

$\to\Delta AHB=\Delta AHC$(cạnh huyền-góc nhọn)

$\to HB=HC$

b.Từ câu a$\to H$ là trung điểm $BC$

$\to HB=HC=\dfrac12BC=4$

Mà $AH\perp BC$

$\to AB^2=AH^2+BH^2$

$\to AH^2=AB^2-BH^2=9$

$\to AH=3$

c.Từ câu a $\to \widehat{BAH}=\widehat{HAC}$

$\to \widehat{DAH}=\widehat{HAE}$

Xét $\Delta AHD,\Delta AHE$ có:

$\widehat{ADH}=\widehat{AEH}(=90^o)$

Chung $AH$

$\widehat{DAH}=\widehat{HAE}$

$\to \Delta AHD=\Delta AHE$(Cạnh huyền-góc nhọn)

$\to HD=HE$

$\to\Delta HDE$ cân tại $H$

d.Ta có $HE\perp AC\to HE<HC$

Mà $HE=HD\to HD<HC$

Bài 4:

a.Xét $\Delta AMB,\Delta AMC$ có:

Chung $AM$

$AB=AC$

$MB=MC$ vì $M$ là trung điểm $BC$

$\to\Delta AMB=\Delta AMC(c.c.c)$

$\to \widehat{MAB}=\widehat{MAC}$

$\to AM$ là phân giác $\hat A$

b.Từ câu a $\to \widehat{AMB}=\widehat{AMC}$

Mà $\widehat{AMB}+\widehat{AMC}=180^o$

$\to \widehat{AMB}=\widehat{AMC}=90^o$

$\to AM\perp BC$

c.Ta có $M$ là trung điểm $BC$

$\to MB=MC=\dfrac12BC=3$

Mà $AM\perp BC$

$\to AM^2=AB^2-BM^2=16$

$\to AM=4$

d.Xét $\Delta AME,\Delta AMF$ có:

$\widehat{AEM}=\widehat{AFM}(=90^o)$

Chung $AM$

$\widehat{EAM}=\widehat{MAF}$ vì $AM$ là phân giác $\hat A$

$\to\Delta AME=\Delta AMF$(cạnh huyền-góc nhọn)

$\to ME=MF$

$\to\Delta MEF$ cân tại $M$

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