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28 tháng 11 2023

Để chứng minh góc BEA vuông, ta cần chứng minh rằng tam giác BEA là tam giác vuông.

 

Ta có các thông tin sau:

- Tam giác ABC cân tại A, do đó góc ABC = góc BAC.

- D là trung điểm của đường cao AH, do đó AD = DH.

- HE vuông góc với DC tại E.

 

Bây giờ, ta sẽ chứng minh tam giác BEA là tam giác vuông bằng cách sử dụng các thông tin trên.

 

Ta có:

- Góc ABC = góc BAC (tam giác ABC cân tại A).

- Góc ABD = góc ADH (hai góc đối nhau).

- AD = DH (D là trung điểm của AH).

 

Vì tam giác ABD và tam giác ADH là hai tam giác đồng dạng (có hai góc bằng nhau và cạnh tương ứng bằng nhau), nên chúng tương đương.

 

Do đó, ta có:

- Góc ADB = góc ADH (tam giác đồng dạng).

- Góc ADB = góc BEA (hai góc đối nhau).

 

Vậy, ta có góc BEA = góc ADH = góc ADB.

 

Vì góc ADB là góc vuông (do AD = DH và HE vuông góc với DC), nên góc BEA cũng là góc vuông.

 

Vậy, ta đã chứng minh được rằng góc BEA là góc vuông.

13 tháng 2 2016

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7 tháng 3 2017

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26 tháng 2 2022

bạn ơi mờ quá

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

AB=AC

AH chung

=>ΔAHB=ΔAHC

=>HB=HC và góc BAH=góc CAH

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

a: Xét ΔAHD và ΔAED có 

AH=AE

\(\widehat{HAD}=\widehat{EAD}\)

AD chung

DO đó: ΔAHD=ΔAED

Suy ra: DH=DE

Ta có: DH=DE

mà DE<DC

nên DH<DC

b: Ta có: AH=AE

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

Ta có: DH=DE

nên D nằm trên đường trung trực của HE(2)

Từ (1) và (2) suy ra AD là đường trung trực của HE

c: \(\widehat{BAD}+\widehat{CAD}=90^0\)

\(\widehat{BDA}+\widehat{HAD}=90^0\)

mà \(\widehat{CAD}=\widehat{HAD}\)

nên \(\widehat{BAD}=\widehat{BDA}\)

hay ΔBDA cân tại B

d: Để ΔBDA đều thì \(\widehat{B}=60^0\)