Median theorem – Wikipedia

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In Euclidean geometry, the median theorem , or Apollonius theorem , designates one of the following three identities [ first ] , on distances and scalar products, in a triangle ABC median WHO and high Like :

First median theorem or Apollonius theorem [ modifier | Modifier and code ]

Apollonius theorem – Be ( ABC ) Any triangle and WHO the median from A . We then have the following relation:

AB2+AC2=2BI2+2AI2{displaystyle AB^{2}+AC^{2}=2BI^{2}+2AI^{2}}

or :

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AB2+AC2=12BC2+2AI2.{displaystyle AB^{2}+AC^{2}={1 over 2}BC^{2}+2AI^{2}.}

This theorem is a reformulation of the identity of the parallelogram.

Demonstration by the scalar product [ modifier | Modifier and code ]

This property is a simple case of reducing the scalar function of Leibniz: just bring the point I in both vectors

AB→{displaystyle {overrightarrow {AB}}}

And

AC→{displaystyle {overrightarrow {AC}}}

, by the Chasles relationship:

A B2+ A C2= ( AI→+ IB→)2+ ( AI→+ IC→)2. {displaystyle AB^{2}+AC^{2}=({overrightarrow {AI}}+{overrightarrow {IB}})^{2}+({overrightarrow {AI}}+{overrightarrow {IC}})^{2}.}

We develop:

A B2+ A C2= A I2+ I B2+ 2 AI→⋅ IB→+ A I2+ I C2+ 2 AI→⋅ IC→. {displaystyle AB^{2}+AC^{2}=AI^{2}+IB^{2}+2{overrightarrow {AI}}cdot {overrightarrow {IB}}+AI^{2}+IC^{2}+2{overrightarrow {AI}}cdot {overrightarrow {IC}}.}

Point I is the middle of [ BC ], SO

IB→{displaystyle {overrightarrow {IB}}}

And

IC→{displaystyle {overrightarrow {IC}}}

are opposed, which implies that scalar products are eliminated and IC 2 = Single 2 SO

A B2+ A C2= 2 A I2+ 2 I B2. {displaystyle AB^{2}+AC^{2}=2AI^{2}+2IB^{2}.}

Demonstration using only theorems on distances [ modifier | Modifier and code ]

Either H the foot of the height from A . The three triangles Ahb , AHC And Raid are rectangles in H ; By applying the Pythagoras theorem to them, we get:

A B2= A H2+ H B2, A C2= A H2+ H C2etA I2= A H2+ H I2. {Displaystyle AB^{2} = AH^{2}+HB^{2}, Quad AC^{2} = AH^{2}+HC^{2} Quad {rm {et}} Quad Ai^{2 } = AH^{2}+hi^{2}.}.

We can deduce :

A B2+ A C2= H B2+ H C2+ 2 A H2= H B2+ H C2+ 2 ( A I2– H I2) . {DisplayStyle ab^{2}+AC^{2} = hb^{2}+HC^{2}+2ah^{2} = hb^{2}+HC^{2} +2 (AI^{2 } -HI^{2}).}

We express HB And HC in terms of HI And WITH A . Even if it means intervening B And C If necessary, we can always assume that B And H are on the same side of I . SO,

H B = |H I – B I | et H C = H I + I C = H I + B I . {Displaystyle hb = | hi-bi | {text {et}} hc = hi+IC = Bi+Bi.}

We can therefore transform, in the expression above of

A B 2+ A C 2{displaystyle AB^{2}+AC^{2}}

, under-expression

HB2+HC2=(HI−BI)2+(HI+BI)2=HI2−2HI⋅BI+BI2+HI2+2HI⋅BI+BI2=2HI2+2BI2.mm Sintle yley toket matal funy ralway like malkal kalkom) hyomom) hyomomhkhééééézézékékékate mize 22-22 25- woeal want to be palant 2alk?

By replacing, we get:

A B2+ A C2= 2 H I2+ 2 B I2+ 2 ( A I2– H I2) = 2 B I2+ 2 A I2. {displaystyle ab^{2}+ac^{2} = 2Hi^{2}+2BI^{2} +2 (AI^{2} -HI^{2}) = 2BI^{2}+2ai^} 2}.}

Generalization to any Cévienne [ modifier | Modifier and code ]

The above demonstration by the scalar product is generalized, which makes it possible to demonstrate:

Be ( ABC ) a triangle, J A point of [ BC ] different from B , And k = JC / JB . SO :

k AB2+AC2=(k+1)(k BJ2+AJ2).{Displaystyle k ~ ab^{2}+AC^{2} = (k+1) (k ~ bj^{2}+aj^{2}).}

Demonstration

Point J is the barium of ( B , k ) And ( C , 1) So ( cf. Scalar function of Leibniz)

k AB2+AC2=k JB2+JC2+(k+1)JA2=k JB2+(k JB)2+(k+1)JA2=(k+k2)JB2+(k+1)JA2=(k+1)(k JB2+JA2).{Displaystyle {begin {aligned} k ~ ab^{2}+AC {2} & = k ~ jb^{2}+jc {2}+(k+1) I^{2} \ & = k = k = k = k = ~ Jb^{2}+(k ~ jb) {2}+(k+1) ja {2} \ & = (k+k^{2}) jb^{2}+(k+1) I^{2} \ & = (k+1) (k ~ jb^{2}+I^{2}). End {alligned}}}

Second median theorem – Be ( ABC ) a triangle and I The middle of the segment [ BC ]. SO

AB→⋅AC→=AI2−14BC2.{displaystyle {overrightarrow {AB}}cdot {overrightarrow {AC}}=AI^{2}-{dfrac {1}{4}}BC^{2}.}

The demonstration uses the same decomposition of the vectors

AB→{displaystyle {overrightarrow {AB}}}

And

AC→{displaystyle {overrightarrow {AC}}}

that above:

AB→⋅ AC→= ( AI→+ IB→) ⋅ ( AI→– IB→) = A I2– I B2= A I2– ( B C /2 )2. {displaystyle {overrightarrow {AB}}cdot {overrightarrow {AC}}=({overrightarrow {AI}}+{overrightarrow {IB}})cdot ({overrightarrow {AI}}-{overrightarrow {IB}})=AI^{2}-IB^{2}=AI^{2}-(BC/2)^{2}.}

Median theorem for a rectangle triangle [ modifier | Modifier and code ]

There is a special case relating to the rectangle triangle.

Theorem – In a rectangle triangle, the length of the median from the top of the right angle is half the length of the hypotenuse.

This theorem has a reciprocal.

Theorem – If in a triangle, the length of the median from a summit is half the length on the opposite side, then this triangle is rectangle on this summit.

Third median theorem – Be ( ABC ) a triangle and I The middle of the segment [ BC ]. On note H The orthogonal projected of A on ( BC ). SO

|AB2−AC2|=2BC×IH.{displaystyle left|AB^{2}-AC^{2}right|=2BCtimes IH.}

More precisely :

A B2– A C2= 2 BC¯⋅ IH¯, {displaystyle AB^{2}-AC^{2}=2~{overline {BC}}cdot {overline {IH}},}

Or BC And THEM designate algebraic measures compared to the same unit director of the right ( BC ).
Just use the scalar product and remarkable identities:

A B2– A C2= ( AB→+ AC→) ⋅ ( AB→– AC→) = 2 AI→⋅ CB→= 2 IA→⋅ BC→. {displaystyle AB^{2}-AC^{2}=({overrightarrow {AB}}+{overrightarrow {AC}})cdot ({overrightarrow {AB}}-{overrightarrow {AC}})=2{overrightarrow {AI}}cdot {overrightarrow {CB}}=2{overrightarrow {IA}}cdot {overrightarrow {BC}}.}

The projection of

IA→{displaystyle {overrightarrow {IA}}}

on ( BC ) East

IH→{displaystyle {overrightarrow {IH}}}

from where

A B2– A C2= 2 IH→⋅ BC→= 2 IH¯⋅ BC¯. {displaystyle AB^{2}-AC^{2}=2~{overrightarrow {IH}}cdot {overrightarrow {BC}}=2~{overline {IH}}cdot {overline {BC}}.}

Stewart Theorem

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