Given △abc,
ad and be are medians, ad ∩be=g.
prove: gd/ag = de/ab

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,

as alternate interior angles.

Moreover, as vertical angles.

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

Hence,

b. ol

step-by-step explanation:

i got the answer right on edg.

congruent: (geometry) (of figures) identical in form; coinciding exactly when superimposed.

alright so i think for this one it would to draw it out. this may be hard for me to try and draw, but i will do my best on here:

        r

  /       |     \

notice how it forms 2 triangles inside one bigger triangle.

so if we know the angle spr is the same as psr that is saying the measure of angle p is the same as angle s. this means the big triangle psr is either isosceles or equilateral. in both cases, the triangle is perfectly symmetrical when split down the middle, which is in this case rq. because we know it is perfectly symmetrical on both sides, we know that the two triangles inside the bigger one are equal.