If Ad, Be And Cf Are The Medians Of A Triangle Abc, Then The True Statement Is:

In triangleABC AD,BE,CF are the medians

neela | Student

Let us have the triangle ABC.

Let D , E and F be the mid points of BC, CA and AB.

Similarly  if we conseder trangle ADC, we get:

(1)+(2):

Similarly we can show by considering triangle BEC and BED that

BC^2+BA^2 = 2BE^2 +(1/2)CA^2.........(5)

Similarly we can show that

CA^2+AB^2 = 2CF^2 +(1/2)AB^2................(4)

(4)+(5)+(6):

Subtract (1/2) (AB^2+BC^2+CA^2) we get:

Multiply by 2:

william1941 | Student

Take the triangle ABC.

We know that AD, BE and CF are the medians. Therefore D , E and F be the mid points of BC, CA and AB resp.

We first consider the triangle ADB:

Now according to the cosine rule :

as D bisects BC

Similarly, for triangle ADC, we get:

=> AB^2 + AC^2 = 2AD^2 + (1/2)*BC^2

Similar results can be obtained for the other medians. Therefore we have :

BC^2 + BA^2 = 2BE^2 +(1/2)*CA^2

CA^2 + AB^2 = 2CF^2 +(1/2)*AB^2

AB^2 + AC^2 = 2AD^2 + (1/2)*BC^2

=> 2(AB^2 + BC^2 + CA^2) = 2(AD^2 + BE^2 + CF^2) + (1/2)*(AB^2+BC^2+CA^2)

=> (3/2)*(AB^2 + BC^2 + CA^2) = 2(AD^2 + BE^2 + CF^2)

Therefore we get

giorgiana1976 | Student

We'll consider the triangle ABC.

We'll apply cosine rule in triangle ADB:

We'll substitute BD = BC/2

We'll apply cosine rule in triangle ADC:

We'll apply cosine rule in triangle BEC and BED:

BC^2+ BA^2 = 2BE^2 + CA^2/2 (5)

We also can prove that:

CA^2+AB^2 = 2CF^2 + AB^2/2(6)