a: Xét ΔCDA vuông tại A và ΔCBA vuông tại A có

CA chug

DA=BA

Do đó:ΔCDA=ΔCBA

b: Ta có: ΔCDB cân tại C

mà CA là đường cao

nên CA là đường phân giác

c: Xét ΔCEI vuông tại E và ΔCFI vuôg tại F có

CI chung

\(\widehat{ECI}=\widehat{FCI}\)

Do đó:ΔCEI=ΔCFI

Suy ra: CE=CF

Xét ΔCDB có CE/CD=CF/CB

nên EF//DB

When it is hot, we often go swimming
What is there in front of your house?
If you does not feel well, you should see the doctor
What color are you baby's eyes?
Hoa learns languages badly
B -> MORE
C -> WATCHES
C -> AT THE MOMENT
C -> OF
C ->IN 
A -> THERE IS
is not a bookshelf in my bedroom
is a small house
is to the right of the bookstore
has many flowers

1. I am not tall enough to reach the ceiling

2. She is sociable enough to make friends easily

3. He is looking at himself in the mirror at the moment 

4. I have to stop smoking

5. Mr. Kerry used to run 5 miles every morning when he was a teenager

6. We really like her sense of humor

7. He ought not to play computer games

8. Hilary doesn't hace to wear uniform at school

1. I am not tall enough to reach the ceiling

2. She is sociable enough to make friends easily

3. He is looking at himself in the mirror at the moment 

4. I have to stop smoking

5. Mr. Kerry used to run 5 miles every morning when he was a teenager

6. We really like her sense of humor

7. He ought not to play computer games

8. Hilary doesn't hace to wear uniform at school

\(1,=\dfrac{x+3-x}{x\left(x+3\right)}=\dfrac{3}{x\left(x+3\right)}\\ 2,=\dfrac{20x+2x-10}{2x-5}=\dfrac{22x-10}{2x-5}\\ 3,=\dfrac{x+5+x-7+x-4}{x-2}=\dfrac{3\left(x-2\right)}{x-2}=3\\ 4,=\dfrac{x^2+3x-x^2-2x-1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\\ 5,=\dfrac{3xy+y^3-3xy-x^3}{x^2y^2}=\dfrac{y^3-x^3}{x^2y^2}\\ 6,=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)

\(7,=\dfrac{x^2+6x-2x+4}{x\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{x\left(x-2\right)}\\ 8,=\dfrac{4x^2+8x+3x-6-32}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2+11x-38}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)\left(4x+19\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x+19}{x+2}\\ 9,=\dfrac{-5x-1-25x^2+15x}{x\left(5x-1\right)\left(5x+1\right)}=\dfrac{-\left(5x-1\right)^2}{x\left(5x-1\right)\left(5x+1\right)}=\dfrac{1-5x}{x\left(5x+1\right)}\\ 10,=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x-4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}=\dfrac{1}{x}\)