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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)
Cách 1 là:
Xét 🔺AHB vuông tại H1 và 🔺AHB vuông tại H2 ,ta có:
AC=AB(vì là tam giác cân)
góc B= góc C(vì là tam giác cân)
=>🔺AHC=🔺AHC cạnh huyền-góc nhọn)
=> H là trung điểm của BC
Cách 2:
Xét 🔺AHC vuông tại H1 và 🔺 AHB vuông tại H2 ,ta có:
AB=AC(vì là tam giác cân)
AH là cạnh chung
=> 🔺AHC=🔺 AHB ( cạnh huyền góc vuông)
=> H là trung điểm của BC
b)
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CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
Xét ΔAHB vuông tại H và ΔAHC vuông tại H có
AB=AC
AH chung
Do đó: ΔAHB=ΔAHC
=>HB=HC
mà H nằm giữa B và C
nên H là trung điểm của BC
Giải
b)Xét tam giác BAH và CAH có:
AB=AC(gt)
góc B =góc C(gt)
AH chung
\(\Rightarrow\)tam giác BAH =CAH (c.g.c)
\(\Rightarrow\)góc BAH=CAH (2 góc t/ư)
Mặt khác AH nằm giữa AB và AC ,chia góc A thành 2 góc bằng nhau
Mà H là trung điểm BC
\(\Rightarrow\)AH là tia phân giác góc A và vuông góc BC
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.
Xét tam giác ABH và ACH có :
AB=AC (gt)
ABC=ACB
BAH=CAH ( vì đ/c cũng là pg)
=> Hai t/g bằng nhau => BH=BC
(P/s: cũng có thể ns là vì AH là đc nên cx là trung tuyến => BH=CH)