Question

ABCD trapez
ABIICD,AB>CD
M apartine AD, N apartine BC
DM/AM = CN/BN
Demonstreaza ca MN ||AB
Am facut si desen.

Answer 1

Răspuns:

Explicație pas cu pas:

Notam {O}=AD∩CB

DO=x

CO=y

DC║AB⇒Thales ⇒$\frac{x}{AD} =\frac{y}{CB}\\\\\frac{AD}{BC} =\frac{x}{y}$

$\frac{DM}{AM} =\frac{CN}{BN}$ facem proportii derivate

$\frac{DM}{AD} =\frac{CN}{BC}\\\\frac{AD}{BC} =\frac{DM}{CN} =\frac{x}{y}$

$\frac{AM}{BN}=\frac{DM}{CN} =\frac{x}{y}=k\\AM=xk=DM\\CN=yk=CN$

Demonstram ca este adevarata relatia

$\frac{AM}{MA} =\frac{AN}{BN} \\\\\\\frac{x+DM}{AM} =\frac{y+CN}{BN}$

Inlocuim

$\frac{x+xk}{xk} =\frac{y+yk}{yk} \\\frac{x(1+k)}{xk} =\frac{y(1+k)}{yk} \\\frac{1}{k} =\frac{1}{k}$⇒Thales ca MN║AB