Question

Prove that medians of a trangle divided each other in the same ratio of 2:1.​

Answer 1

Step-by-step explanation:

Let there be a triangle ABC, with medians AD , BE and CF respectively.

To prove : AO/DO =CO/OF=BO/OE=2/1

Construction: Draw a line from point D to the side AC such that intersect AC at point X and such that DX is parallel to BE.

So, now we start the proof: In triangle BCE, DX is parallel to BE, so CD/BD =CX/XE ( by basic proportionality theorem in which if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio) .

Now BD=CD as AD is a median on side BC. So, CD/CD=CX/XE

1/1=CX/XE

CX=XE (by cross multiplication)

Now in triangle ADC, OE is parallel to DX. So,

AE/EX=AO/DO

Now AE=AC/2 (because BE is a median on side AC)

And, EX=CE/2 ( because we have proved above that CE=EX, so X is the mid point of CE)

Hence, AC/2 /CE/2=AO/DO.

Now, 2 gets cancel so AC/CE=AO/DO

2CE/CE=AO/DO ( as BE is the median on side AC so AC=2CE)

Now, CE gets cancel so,

similarly , we can proof :-

AO/DO=CO/OF=BO/OE=2/1.

Hence, proved

Hope it helps you

Answer 2

Answer:

Step-by-step explanation:

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