Cho tam giác ABC cân tại A. BD,CE là đường cao. AB=c, BC=a, AC=b. Chứng minh rằng: \(DE=\dfrac{a\left(2b^2-a^2\right)}{2b^2}\)
Những câu hỏi liên quan
Cho tam giác ABC vuông tại A có BC a, CA b, AB c, đường cao AH.a) Chứng minh: 1+tam^2Bdfrac{1}{cos^2B};tandfrac{C}{2}dfrac{c}{a+b}b) Chứng minh: AH a. sin B. cos B, BHa·cos2B, CHa·sin2Bc) Lấy D trên cạnh AC. Kẻ DE vuông góc BC tại E. Chứng minh:sinBdfrac{ABcdot AD+EBcdot ED}{ABcdot BE+DAcdot DE} (
Đọc tiếp
Cho tam giác ABC vuông tại A có BC = a, CA = b, AB = c, đường cao AH.
a) Chứng minh: \(1+tam^2B=\dfrac{1}{cos^2B};tan\dfrac{C}{2}=\dfrac{c}{a+b}\)
b) Chứng minh: AH = a. sin B. cos B, BH=a·cos2B, CH=a·sin2B
c) Lấy D trên cạnh AC. Kẻ DE vuông góc BC tại E. Chứng minh:
sinB=\(\dfrac{AB\cdot AD+EB\cdot ED}{AB\cdot BE+DA\cdot DE}\) (
a) \(1+tan^2B=1+\dfrac{AC^2}{AB^2}=\dfrac{AB^2+AC^2}{AB^2}=\dfrac{BC^2}{AB^2}=\dfrac{1}{\left(\dfrac{AB}{BC}\right)^2}=\dfrac{1}{cos^2B}\)
b) Ta có: \(a.sinB.cosB=BC.\dfrac{AC}{BC}.\dfrac{AB}{BC}=\dfrac{AC.AB}{BC}=\dfrac{AH.BC}{BC}=AH\)
\(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=BC.\left(\dfrac{AB}{BC}\right)^2=BC.cos^2B\)
Tương tự \(\Rightarrow CH=BC.sin^2B\)
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1. Cho tam giác ABC vuông ở A có AB<AC. AH vuông góc với BC tại H, D là điểm trên cạnh BC sao cho AD=AB. Vẽ DE vuông góc với BC tại E. Chứng mih rằng AH=HE.
2. Cho tam giác ABC vuông cân tại A.. Qua A vẽ đường thẳng d ở ngoài tam giác ABC . Vẽ BD vuông góc với d taị D. CE vuông góc với d tại E. M là trung điểm CB. Chứng minh rằng:
a) BD + CE = DE
b) Tam giác MDE là tam giác vuông cân
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cho tam giác ABC vuông tại A (AB < AC). Về phía ngoài tam giác ABC vẽ 2 tam giác ABD và tam giác ACE vuông cân ở A
a) CM BC = DE
b)CM BD song song với CE
c)Kẻ dường cao AH của tam giác ABC cắt DE tại M. Vẽ đường thẳng qua A và vuông góc với MC cắt BC tại. Chứng minh rằng CA vuông góc với NM
d) CM rằng AM = 1 phần 2 DE
a) Xét \(\Delta ABC\)và\(\Delta ADE\):
AB=AD(gt)
\(\widehat{BAC}=\widehat{DAE}=90^o\)
AC=AE(gt)
=> \(\Delta ABC=\Delta ADE\left(c-g-c\right)\)
=> BC=DE ( 2 cạnh tương ứng)
=> Đpcm
b) Ta có \(\Delta ABD\)vuông cân tại A
=> \(\widehat{ABD}=\widehat{ADB}=\frac{\widehat{DAB}}{2}=\frac{90^o}{2}=45^o\)
\(\Delta AEC\)vuông cân tại A
=> \(\widehat{AEC}=\widehat{ACE}=\frac{\widehat{EAC}}{2}=\frac{90^o}{2}=45^o\)
=> \(\widehat{BDA}=\widehat{ECA}=45^o\)
Mà 2 góc này ở vị trí so le trong
=> BD//CE
=> Đpcm
c) Sửa đề: Kẻ dường cao AH của tam giác ABC cắt DE tại M. Vẽ đường thẳng qua A và vuông góc với MC cắt BC tại N. Chứng minh rằng CA vuông góc với NM
Gọi giao điể của NA và MC là I
Xét \(\Delta NMC\)có:
\(\hept{\begin{cases}NI\perp MC\\MH\perp NC\end{cases}}\)
Mà 2 đường cao này cắt nhau tại A
=> A là trực tâm của \(\Delta MNC\)
=> \(CA\perp NM\)
=> Đpcm
d) Ta có: \(\widehat{ADM}=\widehat{ABC}\left(\Delta ADE=\Delta ABC\right)\)
=> \(\widehat{ADM}+\widehat{AED}=\widehat{ABC}+\widehat{BAH}=90^o\)
=> \(\widehat{AED}=\widehat{BAH}\) Mà \(\widehat{BAH}=\widehat{MAE}\left(đđ\right)\)
=> \(\widehat{AED}=\widehat{MAE}\)
=> \(\Delta MAE\)cân tại M
=> MA=ME (1)
Lại có: \(\widehat{AED}=\widehat{ACB}\Rightarrow\widehat{AED}+\widehat{ADE}=\widehat{ACB}+\widehat{CAH}=90^o\)
=> \(\widehat{ADE}=\widehat{CAH}\)
Mà \(\widehat{CAH}=\widehat{DAM}\left(đđ\right)\)
=> \(\widehat{ADE}=\widehat{DAM}\)
=> \(\Delta DAM\)cân tại M
=> MD=MA (2)
Từ (1) và (2)
=> MA=MD=ME
=> \(MA=\frac{1}{2}DE\)
=> Đpcm
P/s: Thật ra định làm tắt cho bạn tự suy luận, nhưng sợ bạn ko hiểu nên thoi, mỏi cả tay:>>>
Cho tam giác ABC vuông tại A(AB<AC). Về phía ngoài tam giác ABC vẽ Tam giác ABD và Tam giác ACE cân tại A
a) Chứng minh BC=DE
b) Chứng minh BD//CE
c) Kẻ đường cao AH Của tam giác ABC cắt DE Tại M. Vẽ đường thẳng qua A và vuông góc với MC Cắt BC tại N. Chứng minh rằng CA vuông góc với NM
Cho tam giác ABC vuông tại A (AB AC). Trên tia đối của tia AC lấy điểm D sao cho AD AB. Trên tia đối của tia AB lấy điểm E sao cho AE AC.1) Chứng minh rằng : BC DE.2) Chứng minh rằng : Tam giác ABD vuông cân và BD // CE.3) Vẽ đường cao AH của tam giác ABC, tia AH cắt cạnh DE tại M. Từ A vẽ đường vuông góc với CM tại K, đường thẳng này cắt BC tại N.Chứng minh rằng : MN // AB và AM 1/2 DE.
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Cho tam giác ABC vuông tại A (AB < AC). Trên tia đối của tia AC lấy điểm D sao cho AD = AB. Trên tia đối của tia AB lấy điểm E sao cho AE = AC.
1) Chứng minh rằng : BC = DE.
2) Chứng minh rằng : Tam giác ABD vuông cân và BD // CE.
3) Vẽ đường cao AH của tam giác ABC, tia AH cắt cạnh DE tại M. Từ A vẽ đường vuông góc với CM tại K, đường thẳng này cắt BC tại N.
Chứng minh rằng : MN // AB và AM = 1/2 DE.
1) Xét ΔCAB vuông tại A và ΔEAD vuông tại A có
AB=AD(gt)
AC=AE(gt)
Do đó: ΔCAB=ΔEAD(hai cạnh góc vuông)
Suy ra: BC=DE(hai cạnh tương ứng)
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2) Xét ΔABD có AB=AD(gt)
nên ΔABD cân tại A(Định nghĩa tam giác cân)
Xét ΔABD cân tại A có \(\widehat{BAD}=90^0\)(gt)
nên ΔABD vuông cân tại A(Định nghĩa tam giác vuông cân)
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Cho tam giác ABC vuông tại A có ABAC. Trên tia đối của tia AC lấy điểm D sao cho ADAB. Trên tia đối của tia AB lấy điểm E sao cho AEAC a. So sánh các góc của tam giác ABC. Chứng minh BDBC b. Chứng minh BCDE, tam giác ABC vuông cân và BC//CE c. Kẻ đường cao AH của tam giác ABC, đường cao AH cắt DE tại M. Từ A kẻ đường vuông góc với CM tại K. đường thẳng này cắt BC tại N. Chứng minh rằng MN//AB
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Cho tam giác ABC vuông tại A có AB<AC. Trên tia đối của tia AC lấy điểm D sao cho AD=AB. Trên tia đối của tia AB lấy điểm E sao cho AE=AC
a. So sánh các góc của tam giác ABC. Chứng minh BD<BC
b. Chứng minh BC=DE, tam giác ABC vuông cân và BC//CE
c. Kẻ đường cao AH của tam giác ABC, đường cao AH cắt DE tại M. Từ A kẻ đường vuông góc với CM tại K. đường thẳng này cắt BC tại N. Chứng minh rằng MN//AB
Cho tam giác ABC nhọn (ABAC) nội tiếp đường tròn (O) . Tiếp tuyến tại A của (O) cắt đường thẳng BC tại M. Vẽ đường cao BF của tam giác ABC. Từ F kẻ đường thẳng song song với MA cắt AB tại E. a) chứng minh rằng MA^2MB.MC suy ra MC/MBAC^2/AB^2b) CE cắt BF tại H. Chứng minh tứ giác BEFC nội tiếp, suy ra AH vuông góc BC tại Dc) gọi I là trung điểm BC. Chứng minh bốn điểm E,F,D,I cùng nằm trên một đường trònd) từ H vẽ đường thẳng vuông góc với HI cắt AB,AC theo thứ tứ tại P,Q. Chứng minh H là trung đ...
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Cho tam giác ABC nhọn (AB<AC) nội tiếp đường tròn (O) . Tiếp tuyến tại A của (O) cắt đường thẳng BC tại M. Vẽ đường cao BF của tam giác ABC. Từ F kẻ đường thẳng song song với MA cắt AB tại E.
a) chứng minh rằng MA^2=MB.MC suy ra MC/MB=AC^2/AB^2
b) CE cắt BF tại H. Chứng minh tứ giác BEFC nội tiếp, suy ra AH vuông góc BC tại D
c) gọi I là trung điểm BC. Chứng minh bốn điểm E,F,D,I cùng nằm trên một đường tròn
d) từ H vẽ đường thẳng vuông góc với HI cắt AB,AC theo thứ tứ tại P,Q. Chứng minh H là trung điểm PQ
a: Xét ΔMBA và ΔMAC có
góc MAB=góc MCA
góc M chung
=>ΔMBA đồng dạng với ΔMAC
=>MB/MA=MA/MC
=>MA^2=MB*MC
=>MC/MB=AB^2/AC^2
b: EF//AM
AM vuông góc OA
=>EF vuông góc OA
=>góc AEF+góc OAE=90 độ
=>góc AEF+(180 độ-góc AOB)/2=90 độ
=>góc AEF+90 độ-góc ACB=90 độ
=>gócAEF=góc ACB
=>góc BEF+góc BCF=180 độ
=>BEFC nội tiếp
=>góc BEC=góc BFC=90 độ
Xét ΔABC có
BF,CE là đường cao
BF căt CE tại H
=>H là trực tâm
=>AH vuông góc CB tại D
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Cho tam giác ABC nhọn (ABAC) nội tiếp đường tròn (O) . Tiếp tuyến tại A của (O) cắt đường thẳng BC tại M. Vẽ đường cao BF của tam giác ABC. Từ F kẻ đường thẳng song song với MA cắt AB tại E. a) chứng minh rằng MA^2MB.MC suy ra MC/MBAC^2/AB^2b) CE cắt BF tại H. Chứng minh tứ giác BEFC nội tiếp, suy ra AH vuông góc BC tại Dc) gọi I là trung điểm BC. Chứng minh bốn điểm E,F,D,I cùng nằm trên một đường trònd) từ H vẽ đường thẳng vuông góc với HI cắt AB,AC theo thứ tứ tại P,Q. Chứng minh H là trung đ...
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Cho tam giác ABC nhọn (AB<AC) nội tiếp đường tròn (O) . Tiếp tuyến tại A của (O) cắt đường thẳng BC tại M. Vẽ đường cao BF của tam giác ABC. Từ F kẻ đường thẳng song song với MA cắt AB tại E.
a) chứng minh rằng MA^2=MB.MC suy ra MC/MB=AC^2/AB^2
b) CE cắt BF tại H. Chứng minh tứ giác BEFC nội tiếp, suy ra AH vuông góc BC tại D
c) gọi I là trung điểm BC. Chứng minh bốn điểm E,F,D,I cùng nằm trên một đường tròn
d) từ H vẽ đường thẳng vuông góc với HI cắt AB,AC theo thứ tứ tại P,Q. Chứng minh H là trung điểm PQ
a: Xét ΔMBA và ΔMAC có
góc MAB=góc MCA
góc M chung
=>ΔMBA đồng dạng với ΔMAC
=>MB/MA=MA/MC
=>MA^2=MB*MC
=>MC/MB=AB^2/AC^2
b: EF//AM
AM vuông góc OA
=>EF vuông góc OA
=>góc AEF+góc OAE=90 độ
=>góc AEF+(180 độ-góc AOB)/2=90 độ
=>góc AEF+90 độ-góc ACB=90 độ
=>gócAEF=góc ACB
=>góc BEF+góc BCF=180 độ
=>BEFC nội tiếp
=>góc BEC=góc BFC=90 độ
Xét ΔABC có
BF,CE là đường cao
BF căt CE tại H
=>H là trực tâm
=>AH vuông góc CB tại D
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Cho tam giác ABC vuông tại A (AB<AC).Về phía ngoài tam giác ABC vẽ hai tam giác ABD và tam giác ACE vuông cân ở A.
a,CMinh BC=DE
b,CMinh BD//CE
c,Kẻ đường cao AH của tam giác ABC cắt DE tại M.Vẽ đường thẳng qua A và vuông góc MC cắt BC tại N. Chứng minh rằng CA vuông góc với NM
d,CMinh AM =DE/2