cho tam giác ABC có góc A<90 độ vẽ ra phía ngoài tam giác đó 2 đoạn thẳng AD vuông góc và bằng AB và AE vuông góc và bằng AC
Chứng minh DC=BE và DC vuông góc với BE
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bài 2:
ta có: AB<AC<BC(Vì 3cm<4cm<5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
bài 2:
ta có: AB <AC <BC (Vì 3cm <4cm <5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
HT mik làm giống bạn Dương Mạnh Quyết
* Theo mình thì phần a) Góc A = 90 độ sẽ hợp lý hơn chứ. Vậy nên mình sẽ làm theo cả hai góc A 90 độ và 80 độ nhé ( Nhưng bài của mình phần b) sẽ theo góc A = 90 độ )
a)
Góc A = 80 độ thì sẽ có thể tam giác ABC là tam giác cân, tam giác ⊥ tại B hoặc C, tam giác ABC là tam giác tù hoặc tam giác nhọn
Góc A = 90 độ thì tam giác ABC là tam giác vuông tại A
b)
Theo phần a), ta có: Tam giác ABC cân tại A
=> Góc B = góc C = ( 180 độ - 70 độ ) : 2 = 55 độ
3:
góc C=90-50=40 độ
Xét ΔABC vuông tại A có sin C=AB/BC
=>4/BC=sin40
=>\(BC\simeq6,22\left(cm\right)\)
\(AC=\sqrt{BC^2-AB^2}\simeq4,76\left(cm\right)\)
1:
góc C=90-60=30 độ
Xét ΔABC vuông tại A có
sin B=AC/BC
=>3/BC=sin60
=>\(BC=\dfrac{3}{sin60}=2\sqrt{3}\left(cm\right)\)
=>\(AB=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\left(cm\right)\)
Ta có hình vẽ :
Hình như thiếu đk là AD = AB ; AE = AC ( nếu k có đk này thì k giải đk )
Chứng minh :
a)
Vì AD và AB nằm trên cùng một nửa mặt phẳng bờ AC
⇒ AB nằm giữa AD và AC
\(\Rightarrow\widehat{DAB}+\widehat{BAC}=\widehat{DAC}\) (1)
Vì AE và AC nằm trên cùng một nửa mặt phẳng bờ AB
⇒ AC nằm giữa AE và AB
\(\Rightarrow\widehat{EAC}+\widehat{CAB}=\widehat{EAB}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\widehat{DAC}=\widehat{EAB}\)
Xét △DAC và △BAE có :
DA = BA ( gt )
\(\widehat{DAC}=\widehat{EAB}\left(cmt\right)\)
AC = AE ( gt )
⇒ △DAC = △BAE ( c.g.c )
⇒ DC = BE ( tương ứng )
b) Gọi giao điểm của AC và BE là K
Gọi giao điểm của DC và BE là O
Có △DAC = △BAE ( cmt )
\(\Rightarrow\widehat{DCA}=\widehat{BEA}\) ( tương ứng )
*) \(\widehat{AKE}=\widehat{CKO}\left(\text{đối đỉnh}\right)\)
Có :
\(\widehat{EAK}+\widehat{AKE}+\widehat{KEA}=180^o\left(\text{đ/l tổng 3 góc của 1 t/g}\right)\)
\(\Rightarrow\widehat{AKE}+\widehat{KEA}=180^o-\widehat{EAK}\)
Có:
\(\widehat{OKC}+\widehat{KCO}+\widehat{COK}=180^o\left(\text{đ/l tổng 3 góc của 1 t/g}\right)\)
⇒ \(\widehat{OKC}+\widehat{KCO}=180^o-\widehat{COK}\)
Mà \(\widehat{DCA}=\widehat{BEA}\left(cmt\right)\)
\(\widehat{AKE}=\widehat{CKO}\left(cmt\right)\)
⇒ \(\widehat{EAK}=\widehat{COK}\)
Mà \(\widehat{EAK}=90^o\)
⇒ \(\widehat{COK}=90^o\)
⇒ DC ⊥ BE
thanks