# In triangleABC AD,BE,CF are the medians Prove that 3(AB^2+BC^2+AC^2)=4(AD^2+BE^2+CF^2)

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Let us have the triangle ABC.

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

Consider the triangle ADB:

AB^2 = AD^2+ BD^2 - 2AD*DB cos ADB . Due to cosine rule.

AB^2 = AD^2 +(BC/2)^2 - AD*BC* cos(ADB)....(1), as D is mid point of BC.

Similarly if we conseder trangle ADC, we get:

AC^2 =AD^2+(1/2 B/2)^2 - AD*BC*cos(ADC)......(2)

(1)+(2):

AB^2+BC^2 = 2AD^2 + ((BC)^2)/2 - AD*BC{cosADB+cosADC).....(3).

But the angles ADB and ADC are supplementary angles. So cosADB +cosADC = 0. Therefore eq (3) becomes:

AB^2+AC^2 = 2AD^2 +(1/2)BC^2.... (4)

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):

2(AB^2+BC^2+CA^2) =2(AD^2+BE^2+CF^2) +(1/2) {AB^2+BC^2+CA^2).

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

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

Multiply by 2:

3(AB^2+BC^2+CA^2) = 4(AD^2+BE^2+CF^2

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 :

AB^2 = AD^2+ BD^2 - 2AD*DB cos ADB .

as D bisects BC

=> AB^2 = AD^2 +(BC/2)^2 - 2*AD*(BC/2)*cos ADB ...(1)

Similarly, for triangle ADC, we get:

AC^2 = AD^2 + DC^2 - AD*DC*cos ADC

=> AC^2 = AD^2 + (BC/2)^2 - 2*AD*(BC/2)*cos ADC ...(2)

Adding (1) and (2)

=> AB^2 + BC^2 = 2AD^2 + 2*(BC/2)^2) - AD*BC*[cos ADB + cos ADC]

Now as ADB and ADC are are supplementary angles. So cos ADB + cos ADC = 0.

=> 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

Adding the three

=> 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)

=> 3(AB^2+BC^2+CA^2) = 4(AD^2+BE^2+CF^2)

**Therefore we get **

**3(AB^2+BC^2+CA^2) = 4(AD^2+BE^2+CF^2)**

We'll consider the triangle ABC.

We'll apply cosine rule in triangle ADB:

AB^2 = AD^2+ BD^2 - 2AD*DB cos ADB

We'll substitute BD = BC/2

AB^2 = AD^2 +(BC/2)^2 - AD*BC* cos(ADB) (1)

We'll apply cosine rule in triangle ADC:

AC^2 =AD^2+(BC/2)^2 - AD*BC*cos(ADC) (2)

We'll add (1)+(2):

AB^2+BC^2 = 2AD^2 + (BC)^2/2 - AD*BC(cosADB+cosADC)(3)

ADB + ADC = 180

cosADB +cosADC = 0

AB^2+AC^2 = 2AD^2 + BC^2/2(4)

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)

We'll add (4)+(5)+(6):

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

We'll subtract (1/2) (AB^2+BC^2+CA^2) both sides:

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

We'll multiply by 2:

**3(AB^2+BC^2+CA^2) = 4(AD^2+BE^2+CF^2) q.e.d.**