Question

The figure shows triangle ABC with medians AF, BD, and CE. Segment AF is extended to H in such a way that segment GH is congruent to segment AG.
Which conclusion can be made based on the given conditions?
Select one:
a. Segment GD is congruent to segment GF.
b. Segment GD is parallel to segment HC.
c. Segment GF is parallel to segment EB.
d. Segment BH is congruent to segment HC.

Answer 1

Answer:

(B)

Step-by-step explanation:

In ΔAHC

AG=GH (given)                                          (1)

As BD is the median of ΔABC bisecting AC at D, therefore

AD=DC                                                       (2)

Thus, From equation (1) and (2), we get

$\frac{AD}{DC}=\frac{AG}{GH}=\frac{1}{1}$

which implies that the segment GD is dividing the sides AH and AC  in same ratio.

Therefore by the converse of Basic proportionality theorem,segment GD is parallel to segment HC.

Hence, option B is correct.

Answer 2

Answer: Segment GD is parallel to segment HC.

Step-by-step explanation:

In ΔAHC

AG=GH (given)...........1

As BD is the median of ΔABC bisecting AC at D

∴AD=DC.........2

⇒$\frac{AD}{DC}=\frac{AG}{GH}=1$(from 1 and 2)

⇒segment GD is dividing the sides AH and AC  in same ratio.

Therefore by the converse of Basic proportionality theorem,segment GD is parallel to segment HC.

Converse of basic probability theorem states that if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.