a: Ta có: ΔABC cân tại A

mà AD là đường trung tuyến

nên D là trung điểm của BC

b: Ta có: ΔADC cân tại D

mà DE là đường cao

nên E là trung điểm của AC

TA có: ΔADC vuông tại D

mà DE là đường trung tuyến

nên DE=AC/2=AE

=>ΔEAD cân tại E

mà \(\widehat{DEA}=90^0\)

nên ΔEAD vuông cân tại E

Hình nháp thôi em .

Ta có : \(\Delta ABC\) cân tại A 

\(\Rightarrow\) góc ABC \(=\) góc ACB 

Ta có : D là trung điểm của BC

\(\Rightarrow DB=DC\)

Xét \(\Delta BDE\) và \(\Delta CDF\) lần lượt vuông tại E và F có :

                góc ABC \(=\) góc ACB (cmt)

              \(DB=DC\left(cmt\right)\)

Do đó : \(\Delta BDE=\Delta CDF\left(ch-gn\right)\)

\(\Rightarrow DE=DF\)

\(\Rightarrow\Delta DEF\) cân tại D

cíu mình với

A B C D E F

a)Xét \(\Delta ABD\) và \(\Delta ACD\) có :

    \(BD=DC\)

     \(\widehat{ABD}=\widehat{ACD}\left(\Delta ABCcân\right)\)

     AB= AC

=>  \(\Delta ABD\) = \(\Delta ACD\) (c-g-c)

b) Vì \(\Delta ABC\) cân tại A nên AD vừa là đường trung tuyến vừa là đường cao

=> \(AD\perp BC\)

*Nếu chx học cách trên thì bạn xem cách dưới đây"

Vì  \(\Delta ABD\) = \(\Delta ACD\) nên \(\widehat{ADB}=\widehat{ADC}\)

mà \(\widehat{ADB}+\widehat{ADC}=180^o\)

=> \(\widehat{ADB}=\widehat{ADC}=\dfrac{180^o}{2}=90^o\)

=> \(AD\perp BC\)

c)Xét \(\Delta EBD\) vuông tại E và \(\Delta FCD\) vuông tại F có :

\(\widehat{EBD}=\widehat{FCD}\)

\(BD=CD\)

=> \(\Delta EBD=\Delta FCD\left(ch-gn\right)\)

d) Vì D là trung điểm của BC nên  \(DC=\dfrac{BC}{2}=\dfrac{12}{2}=6cm\)

Xét \(\Delta ADC\) vuông tại D có :

\(AC^2=AD^2+DC^2\)

\(100=AD^2+36\)

\(AD^2=100-36\)

\(AD^2=64\)

AD=8 cm

uôi dài v**

ủa r viết ngần đó thì mất bn tg thek

Hình vẽ chuẩn và lời giải hay nhất

a)Vì tam giác abc cân ở a =>góc abc=góc acb.mà góc acb =góc ecn (đối đỉnh) =>góc abc=góc ecn.

Xét tam giác bmd và tam giác cne có :bd=ce; góc abc=góc ecn =>tam giác bmd =tam giác ecn(cạnh góc vuông và góc nhọn kề)

=>md=ne.

b)Vì dm và en cung vuông góc với bc =>dm song song với en=>góc dmc=góc enc(so le trong)

xét tam giác dim và tam giác ein có :góc dmc =góc enc;góc mid=góc nie(đối đỉnh);góc mdi=góc nei=90 độ=>tam giác dim=tam giác ein(g.g.g.)

=>di=ie=>i là trung điểm de

c)gọi h là giao của ao với bc.

ta có:xét tam giác abo bằng tam giác aco=>bo=co=>o thuộc trung trực của bc .tương tự a thuộc trung trực của bc=>ao là trung trực bc

a: Xét ΔMBD vuông tại D và ΔNCE vuông tại E co

MB=NC

góc MBD=góc NCE
=>ΔMBD=ΔNCE

=>MD=NE

b: Xet tứ giác MDNE có

MD//NE

MD=NE

=>MDNE là hình bình hành

=>MN cắt DE tại trung điểm của mỗi đường

=>I là trung điểm của DE

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