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a: Xét ΔAHB vuông tại H và ΔAHC vuông tại H có

AB=AC

AH chung

=>ΔAHB=ΔAHC

=>HB=HC

=>H là trung điểm của BC

b: Xét ΔAMH vuông tại M và ΔANH vuông tại N có

AH chung

góc MAH=góc NAH

=>ΔAMH=ΔANH

=>AM=AN

=>ΔAMN cân tại A

29 tháng 2 2020

Tự vẽ hình nhé ?
a) Vì tam giác ABC cân tại A (GT)
=> Góc ABC = ACB (định lý) (1)
Vì tam giác ABC cân tại A (GT)
=> AB = AC (định nghĩa) (2)
Xét tam giác ABD và tam giác ACD có:
Góc ADB = ADC = 90o (Vì AD vuông góc BC (GT))
AB = AC (Theo (2))
Góc ABC = ACB (Theo (1))
=> Tam giác ABD = tam giác ACD (cạnh huyền - góc nhọn) (3)
=> BD = CD (2 cạnh t.ứng)
Mà D nằm giữa B và C
=> D là trung điểm của BC (đpcm) 
b) Từ (3) => Góc BAD = CAD (2 góc t.ứng) (4)
Mà AD nằm giữa AB và AC
=> AD là tia pg của góc BAC (đpcm)
c) Xét tam giác AED và tam giác AFD có :
Góc AED = AFD = 90o (Vì DE vuông góc AB, DF vuông góc AC (GT))
AD chung
Góc BAD = CAD (Theo (4))
=> Tam giác AED = tam giác AFD (cạnh huyền - góc nhọn)
=> ED = FD (2 cạnh t.ứng)
Xét tam giác DEF có ED = FD (cmt)
=> Tam giác DEF cân tại D (định nghĩa)
Vậy ...

29 tháng 2 2020

bạn ơi đpcm là gì á?

4 tháng 2 2022

a) Xét △AHB và △AHC có:

AB = AC (gt)

BH = HC (gt)

AH Chung

=>△AHB = △AHC (c.c.c)

Do đó góc A1 = góc A2 (2 góc tương ứng)

Mà H là trung điểm của BC => AH vuông góc với BC

b) Xét △AHM và △AHN có:

Góc A1 = Góc A2 (cmt)

Góc M = Góc N (gt)

AH Chung

=> △AHM = △AHN (Cạnh huyền - Góc nhọn)

c) Vì △AHM = △AHN (cmt)

=> AM = AN (2 cạnh tương ứng)

Vì I là giao điểm của MH và AC, K là giao điểm của NH và AB.

=>AK = AI

Do đó: △AIK là tam giác cân (Do có 2 cạnh bằng nhau)

4 tháng 2 2022

tham khảo đâu

13 tháng 2 2016

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7 tháng 3 2017

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a: AH=8cm

b: Xét ΔABH vuông tại H và ΔACH vuông tại H có

AB=AC

AH chung

Do đó: ΔABH=ΔACH

c: Xét ΔDBH và ΔECH

DB=EC

\(\widehat{B}=\widehat{C}\)

BH=CH

Do đó: ΔDBH=ΔECH

Suy ra: HD=HE

hay ΔHDE cân tại H

d: Ta có: AD=AE

nên A nằm trên đường trung trực của DE(1)

Ta có: HD=HE

nên H nằm trên đường trung trực của DE(2)

Từ (1) và (2) suy ra AH là đường trung trực của DE

27 tháng 5 2016

a)Xet 2 tam giác ADF va BDE có BD=AD                                                                                                                                                                                                   goc ADF=goc BDE                                                                                                                                                                                      DF=DE                                                                                                                                                                                  => tam giac ADF=tam giac BDE                                                                                                                                                                 => goc AFD= goc BFD                                                                                                                                                                                 => goc AFD=90                                                                                                                                                                                         AF vuong goc voi FE                                                                                                                                                                

27 tháng 5 2016

a) Xét 2 tam giác ADF và BDE có: BD=AD                                                                                                                                                                                                    góc ADF=góc BDE

17 tháng 12 2021

undefined

Bài 5: 

a) Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=5^2+12^2=169\)

hay BC=13(cm)

Vậy: BC=13cm

b) Xét ΔABE vuông tại B và ΔDBE vuông tại B có 

EB chung

BA=BD(B là trung điểm của AD)

Do đó: ΔABE=ΔDBE(hai cạnh góc vuông)

Suy ra: EA=ED(Hai cạnh tương ứng)

Xét ΔEAD có EA=ED(cmt)

nên ΔEAD cân tại E(Định nghĩa tam giác cân)

6 tháng 5 2021

Cái định lý pytago ấy kq = 179 mà sao lại = 169?

 

a: \(BC=\sqrt{6^2+8^2}=10\left(cm\right)\)

b: Xét ΔBCD có 
BA là đường cao

BA là đường trung tuyến

Do đó: ΔBCD cân tại B