pls solve 22 question
Answers
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
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 :
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)