[{"@context":"http:\/\/schema.org\/","@type":"BlogPosting","@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/median-theorem-wikipedia\/#BlogPosting","mainEntityOfPage":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/median-theorem-wikipedia\/","headline":"Median theorem – Wikipedia","name":"Median theorem – Wikipedia","description":"before-content-x4 A wikipedia article, free l’encyclop\u00e9i. after-content-x4 In Euclidean geometry, the median theorem , or Apollonius theorem , designates one","datePublished":"2017-08-26","dateModified":"2017-08-26","author":{"@type":"Person","@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/author\/lordneo\/#Person","name":"lordneo","url":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/author\/lordneo\/","image":{"@type":"ImageObject","@id":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","url":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","height":96,"width":96}},"publisher":{"@type":"Organization","name":"Enzyklop\u00e4die","logo":{"@type":"ImageObject","@id":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","url":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","width":600,"height":60}},"image":{"@type":"ImageObject","@id":"https:\/\/upload.wikimedia.org\/wikipedia\/commons\/thumb\/b\/b4\/Mediane.svg\/220px-Mediane.svg.png","url":"https:\/\/upload.wikimedia.org\/wikipedia\/commons\/thumb\/b\/b4\/Mediane.svg\/220px-Mediane.svg.png","height":"220","width":"220"},"url":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/median-theorem-wikipedia\/","wordCount":7563,"articleBody":"     (adsbygoogle = window.adsbygoogle || []).push({});before-content-x4A wikipedia article, free l’encyclop\u00e9i.         (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4In 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 :         (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4AB2+AC2=12BC2+2AI2,AB\u2192\u22c5AC\u2192=AI2\u221214BC2,|AB2\u2212AC2|=2BC\u00d7IH.{displaystyle {begin{aligned}AB^{2}+AC^{2}&={frac {1}{2}}BC^{2}+2AI^{2},\\{overrightarrow {AB}}cdot {overrightarrow {AC}}&=AI^{2}-{frac {1}{4}}BC^{2},\\left|AB^{2}-AC^{2}right|&=2BCtimes IH.end{aligned}}}  Table of ContentsFirst median theorem or Apollonius theorem [ modifier | Modifier and code ] Demonstration by the scalar product [ modifier | Modifier and code ] Demonstration using only theorems on distances [ modifier | Modifier and code ] Generalization to any C\u00e9vienne [ modifier | Modifier and code ] Median theorem for a rectangle triangle [ modifier | Modifier and code ] 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 :        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4AB2+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\u2192{displaystyle {overrightarrow {AB}}} And AC\u2192{displaystyle {overrightarrow {AC}}} , by the Chasles relationship: A B2+ A C2= ( AI\u2192+ IB\u2192)2+ ( AI\u2192+ IC\u2192)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\u2192\u22c5 IB\u2192+ A I2+ I C2+ 2 AI\u2192\u22c5 IC\u2192. {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\u2192{displaystyle {overrightarrow {IB}}} And IC\u2192{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 |\u00a0et\u00a0H 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\u2212BI)2+(HI+BI)2=HI2\u22122HI\u22c5BI+BI2+HI2+2HI\u22c5BI+BI2=2HI2+2BI2.mm Sintle yley toket matal funy ralway like malkal kalkom) hyomom) hyomomhkh\u00e9\u00e9\u00e9\u00e9\u00e9z\u00e9z\u00e9k\u00e9k\u00e9kate 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 C\u00e9vienne [ 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\u00a0AB2+AC2=(k+1)(k\u00a0BJ2+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\u00a0AB2+AC2=k\u00a0JB2+JC2+(k+1)JA2=k\u00a0JB2+(k\u00a0JB)2+(k+1)JA2=(k+k2)JB2+(k+1)JA2=(k+1)(k\u00a0JB2+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\u2192\u22c5AC\u2192=AI2\u221214BC2.{displaystyle {overrightarrow {AB}}cdot {overrightarrow {AC}}=AI^{2}-{dfrac {1}{4}}BC^{2}.}  The demonstration uses the same decomposition of the vectors AB\u2192{displaystyle {overrightarrow {AB}}} And AC\u2192{displaystyle {overrightarrow {AC}}} that above: AB\u2192\u22c5 AC\u2192= ( AI\u2192+ IB\u2192) \u22c5 ( AI\u2192– IB\u2192) = 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\u2212AC2|=2BC\u00d7IH.{displaystyle left|AB^{2}-AC^{2}right|=2BCtimes IH.}  More precisely : A B2– A C2= 2  BC\u00af\u22c5 IH\u00af, {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\u2192+ AC\u2192) \u22c5 ( AB\u2192– AC\u2192) = 2 AI\u2192\u22c5 CB\u2192= 2 IA\u2192\u22c5 BC\u2192. {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\u2192{displaystyle {overrightarrow {IA}}} on ( BC ) East IH\u2192{displaystyle {overrightarrow {IH}}} from where A B2– A C2= 2  IH\u2192\u22c5 BC\u2192= 2  IH\u00af\u22c5 BC\u00af. {displaystyle AB^{2}-AC^{2}=2~{overrightarrow {IH}}cdot {overrightarrow {BC}}=2~{overline {IH}}cdot {overline {BC}}.}  Stewart Theorem        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4"},{"@context":"http:\/\/schema.org\/","@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"item":{"@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/#breadcrumbitem","name":"Enzyklop\u00e4die"}},{"@type":"ListItem","position":2,"item":{"@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/median-theorem-wikipedia\/#breadcrumbitem","name":"Median theorem – Wikipedia"}}]}]