Bài 3:
a: Xét hình thang ABCD có MN//AB//CD
nên \(\dfrac{AM}{MD}=\dfrac{BN}{NC}\)
b:
Ta có: \(\dfrac{AM}{MD}=\dfrac{BN}{NC}\)
=>\(\dfrac{MD}{AM}=\dfrac{CN}{NB}\)
=>\(\dfrac{MD+AM}{AM}=\dfrac{CN+NB}{NB}\)
=>\(\dfrac{AD}{AM}=\dfrac{CB}{NB}\)
=>\(\dfrac{AM}{AD}=\dfrac{BN}{BC}\)
\(\dfrac{AM}{AD}+\dfrac{CN}{CB}\)
\(=\dfrac{AM}{AD}+\dfrac{CB-BN}{CB}\)
\(=\dfrac{AM}{AD}+1-\dfrac{BN}{BC}\)
\(=1+\dfrac{AM}{AD}-\dfrac{AM}{AD}=1\)
Bài 2:
a: Xét ΔABC có MN//BC
nên \(\dfrac{AM}{AB}=\dfrac{AN}{AC}\)
=>\(\dfrac{4}{AB}=\dfrac{5}{8,5}=\dfrac{10}{17}\)
=>\(AB=17\cdot\dfrac{4}{10}=\dfrac{68}{10}=6,8\)
AM+MB=AB
=>x+4=6,8
=>x=2,8
b:
ta có: DQ+QF=DF
=>DQ+9=24
=>DQ=15
Xét ΔDEF có PQ//EF
nên \(\dfrac{DP}{PQ}=\dfrac{DQ}{DF}\)
=>\(\dfrac{x}{10,5}=\dfrac{15}{9}=\dfrac{5}{3}\)
=>\(x=10.5\cdot\dfrac{5}{3}=3.5\cdot5=17,5\)
