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Question Number 140735 by ZiYangLee last updated on 12/May/21

Using vector method, prove that three median of a triangle are concurrent.

Usingvectormethod,provethatthreemedianofatriangleareconcurrent.

Answered by MJS_new last updated on 12/May/21

let A= ((0),(0) ) ∧B= ((c),(0) ) ∧C= ((p),(h) ) M_a =((B+C)/2)= ((((c+p)/2)),((h/2)) ) M_b =((A+C)/2)= (((p/2)),((h/2)) ) M_c =((A+B)/2)= (((c/2)),(0) ) v_a =C−B= (((p−c)),(h) ) v_b =A−C= (((−p)),((−h)) ) v_c =B−A= ((c),(0) ) n_a = (((−h)),((p−c)) ) n_b = ((h),((−p)) ) n_c = ((0),(c) ) medians m_a : X=M_a +λ_a n_a ⇔ y=((c−p)/h)x−((c^2 −h^2 −p^2 )/(2h)) (I) m_b : X=M_b +λ_b n_b ⇔ y=−(p/h)x+((h^2 +p^2 )/(2h)) (II) m_c : X=M_c +λ_c n_c ⇔ x=(c/2) inserting x=(c/2) in both equations I and II we get the same value for y y=((−cp+h^2 +p^2 )/(2h)) ⇒ m_a ∩m_b ∩m_c = (((c/2)),(((−cp+h^2 +p^2 )/(2h))) )

letA=(00)∧B=(c0)∧C=(ph)Ma=B+C2=(c+p2h2)Mb=A+C2=(p2h2)Mc=A+B2=(c20)va=C−B=(p−ch)vb=A−C=(−p−h)vc=B−A=(c0)na=(−hp−c)nb=(h−p)nc=(0c)mediansma:X=Ma+λana⇔y=c−phx−c2−h2−p22h(I)mb:X=Mb+λbnb⇔y=−phx+h2+p22h(II)mc:X=Mc+λcnc⇔x=c2insertingx=c2inbothequationsIandIIwegetthesamevalueforyy=−cp+h2+p22h⇒ma∩mb∩mc=(c2−cp+h2+p22h)

Answered by peter frank last updated on 12/May/21

Commented by peter frank last updated on 12/May/21

The median of triangle cuts another median in ratio (2:1) from AD→E(((B^→ +C^→ )/2)) P=((1×A+((2(B^→ +C^→ ))/2))/3)=((A^→ +B^→ +C^→ )/3) from BE→(((A^→ +C^→ )/2)) P=((1×B+((2(A^→ +C^→ ))/2))/3)=((A^→ +B^→ +C^→ )/3) from FC→(((A^→ +B^→ )/2)) P=((1×C+((2(A^→ +B^→ ))/2))/3)=((A^→ +B^→ +C^→ )/3) hence its concurent

Themedianoftrianglecutsanothermedianinratio(2:1)fromAD→E(B→+C→2)P=1×A+2(B→+C→)23=A→+B→+C→3fromBE→(A→+C→2)P=1×B+2(A→+C→)23=A→+B→+C→3fromFC→(A→+B→2)P=1×C+2(A→+B→)23=A→+B→+C→3henceitsconcurent

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