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frika

Answer:

See explanantion

Step-by-step explanation:

Since AD and BE are medians, then points D and E are midpoints of the sides BC and AC, respectively. Thus, segment DE is triangle's midline. By the Triangle midline theorem, DE is parallel to the side AB. Thus,

  • [tex]\angle ABE\cong \angle BED;[/tex]
  • [tex]\angle BAD\cong \angle EDA[/tex]

as alternate interior angles.

Moreover, [tex]\angle BGA\cong \angle EGD[/tex] as vertical angles.

This gives us that triangles ABG and DEG are similar by AAA postulate.

Hence,

[tex]\dfrac{AB}{ED}=\dfrac{AG}{GD}[/tex]

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MrRoyal

The triangle median bisect each side of the triangle

The statement [tex]\frac{GD }{ AG} = \frac{DE }{ AB}[/tex] is true

The given parameters are:

  1. Triangle ABC
  2. Medians: AD and BE
  3. G = AD ∩ BE

To prove the statement, we make use of the Triangle midline theorem.

The theorem states that:

A triangle cut along its midline creates a parallel segment to the base of the triangle and half as long as the base.

In this case, line segment DE is parallel to line segment AB.

By alternate angles, we have:

[tex]\angle ABE \cong \angle BED[/tex]

[tex]\angle BAD \cong \angle EDA[/tex]

And by vertical angles, we have:

[tex]\angle BGA \cong \angle EGD[/tex]

This means that triangles DEG and ABG are similar triangles by the AAA postulate.

So, we have the following equivalent ratios

[tex]GD : AG = DE : AB[/tex]

Express as fractions

[tex]\frac{GD }{ AG} = \frac{DE }{ AB}[/tex]

Hence, the statement has been proved

Read more about triangle medians at:

https://brainly.com/question/7856454

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