Medians of a triangl

Question

Medians AX and BY of Triangle ABC are perpendicular at point G. Prove that AB=CG.

Answer 1

Extend the BY to point D that point D is from drawing a vertical line CD for CD⊥BY(or we say BD) at point D. (point D is out side the triangle)

 

STEP 1:

 

In the ΔBCD,

 

1) XG ⊥ BD  and  CD ⊥ BD  , so XG and CD are parallel

2) BX =1/2 BC (Because AX is Median)

 

From that two condition we can get that BG = 1/2 BD, which means BG=GD,

 

 

STEP 2:

 

Then let's see the ΔAGY and ΔCDY, 

1)They have the vertically opposite angles ∠AYG and ∠CYD, which means angle ∠AYG = ∠CYD;

2)Both of them has a right angle ∠AYD=∠CYD = 90°

3)They have tow equal side that AY=CY, because BY is the Median of ΔABC

 

Using the three conditions above, we can get that ΔAGY and ΔCDY are congruent triangles, so we can say AG=CD because they are the corresponding sides of two congruent triangles ΔAGY and ΔCDY

 

STEP 3:

 

Now from the two steps above, we get conclusions as following:

 

1) BG=GD

2) AG=CD

3) AX ⊥ BD and CD ⊥ BD which means ∠AGB = ∠CDG = 90°

 

So we can know that ΔAGB and ΔCDG are congruent triangles, so the corresponding sides AB=CG

 

Or in other hand, because of they are right angle triangles, we can use the Pythagorean theorem as:

 

AB²=AG²+BG²

CG²=CD²+GD²

 

Because BG=GD and AG=CD, we will see that AB=CG.