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3 years ago

Given △ABC, AD and BE are medians, AD ∩BE=G. Prove: GD/AG = DE/AB

Mathematics

1 answer:

saw5 [17]3 years ago

3 0

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,

$\angle ABE\cong \angle BED;$
$\angle BAD\cong \angle EDA$

as alternate interior angles.

Moreover, $\angle BGA\cong \angle EGD$ as vertical angles.

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

Hence,

$\dfrac{AB}{ED}=\dfrac{AG}{GD}$

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