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2 tháng 5 2017

bạn nào giúp mk vẽ hình đc không

27 tháng 2 2020

Xét ΔADE và ΔABC có :
AD = AB (gt)

góc DAE =góc BAC = 90 độ
AE = AC (gt)
Do đó : ΔADE = ΔABC(c − g − c)
⇒ DE = BC ( hai cạnh tương ứng )
b.
Ta có :
góc ADE =góc CDN ( hai góc đối đỉnh )
góc C= góc E
( vì ΔADE = ΔABC )
⇒ góc N = góc A 90đọ
Hay DE ⊥ BC
Vậy DE ⊥ BC

a: Xét ΔABC vuông tại A và ΔADE vuông tại A có

AB=AD

AC=AE

Do đó: ΔABC=ΔADE

b: Xét ΔAMD và ΔANB có

AM=AN

MD=NB

AD=AB

Do đó: ΔAMD=ΔANB

16 tháng 7 2021

Câu C bạn cm AFCE là hình chữ nhật , FE là đường chéo => E,F,M thẳng hàng vì 2 đường chéo hình chữ nhật đi qua trung điểm của mỗi đường.

 

18 tháng 4 2018

hình các bn tự vẽ nhé(mog các bn thông cảm máy mk ko vẽ dc hình)

a,             Xét tam giác BDA và tam giác MDA,có

                      AD cạnh chung

                   góc BAD=góc MAD (vì AD là tia phân giác của góc A)

                   BA=MA(gt)

            Do đó tam giác BDA= tam giác MDA(c-g-c)

   Suy ra BD=MD(2 cạnh tương ứng)

b,

TA có :góc ABD+góc DBE= 180 độ

           góc AMD + góc DMC =180 độ

Mà góc ABD= góc AMD (cmt)

suy ra góc DBE= góc DMC

                  Xét tam giác BDE và tam giác MDC ,có:

                                góc BDE=góc MDC(2 góc đối đỉnh)

                              BD=MD(cmt)

                              góc  DBE= góc DMC(cmt)

                   Do đó tam giác BDE =tam giác MDC (g-c-c)

s c,d mk đang nghĩ chưa ra kết quả khi nào ra mk giải tiếp heheh thông cảm

18 tháng 4 2018

ko biết

sorry , I don 't no

Kb nhé

A B C M

Ta có : AB = AC => tam giác ABC cân tại A

Ta lại có :

 B = C ( do ABC cân )

AH chung

BM = MC ( gt )

=> AMB = AMC ( c- g - c )

b) Ta có ABC cân 

MÀ M là trung điểm của BC

=> AM là đường cao của ABC

=> AM vuông với BC

13 tháng 7 2016

A B C D E M .. ..

a)  Xét \(\Delta AMB\)và \(\Delta AMC\)có:

AB = AC (gt)

AM : cạnh chung (gt)

BM = CM (gt)

\(\Rightarrow\Delta AMB=\Delta AMC\left(c.c.c\right)\)

b) \(\Delta ABC\): có M là trung điểm BC => AM  là đường trụng trực của BC.

Mà \(\Delta ABC\)cân tại A nên đường trụng trực đồng thời cũng là đường cao. 

\(\Rightarrow AM\)vuông góc \(BC\)

c) Xét \(\Delta ABE\)và \(\Delta ACD\)có:

AC = AB  (gt)>
Góc A : góc chung (gt)

Do AB = AC(gt) : BD = CE (gt)

=> AB - BD = AC - CE 

=> AD = AE.

Vậy \(\Delta ABE=\Delta ADC\)(c.g.c)

d) \(\Delta ABC\)cân có:

BD = CE

2 đoạn thằng cách đều BC nên khi kẻ DE thì \(DE\)//\(BC\).

13 tháng 2 2016

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7 tháng 3 2017

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