Difference between revisions of "Viviani's theorem"
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We label the altitudes from <math>P</math> to each of sides <math>\overline{AB}</math>, <math>\overline{BC}</math> and <math>\overline{AC}</math> <math>x</math>, <math>y</math> and <math>z</math> respectively. Since <math>\triangle ABC</math> is equilateral, we can say that <math>s=AB=BC=AC</math>. Therefore, <math>[ABP]=\dfrac{sx}{2}</math>, <math>[BCP]=\dfrac{sy}{2}</math> and <math>[ACP]=\dfrac{sz}{2}</math>. Since the area of a triangle is the product of its base and altitude, we also have <math>[ABC]=\dfrac{as}{2}</math>. However, the area of <math>\triangle ABC</math> can also be expressed as <math>[ABC]=[ABP]+[BCP]+[ACP]=\dfrac{sx}{2}+\dfrac{sy}{2}+\dfrac{sz}{2}=\dfrac{s}{2}(x+y+z)</math>. Therefore, <math>\dfrac{s}{2}(x+y+z)=\dfrac{s}{2}(a)</math>, so <math>x+y+z=a</math>, which means the sum of the altitudes from any point within the triangle is equal to the altitude from the vertex of a triangle. | We label the altitudes from <math>P</math> to each of sides <math>\overline{AB}</math>, <math>\overline{BC}</math> and <math>\overline{AC}</math> <math>x</math>, <math>y</math> and <math>z</math> respectively. Since <math>\triangle ABC</math> is equilateral, we can say that <math>s=AB=BC=AC</math>. Therefore, <math>[ABP]=\dfrac{sx}{2}</math>, <math>[BCP]=\dfrac{sy}{2}</math> and <math>[ACP]=\dfrac{sz}{2}</math>. Since the area of a triangle is the product of its base and altitude, we also have <math>[ABC]=\dfrac{as}{2}</math>. However, the area of <math>\triangle ABC</math> can also be expressed as <math>[ABC]=[ABP]+[BCP]+[ACP]=\dfrac{sx}{2}+\dfrac{sy}{2}+\dfrac{sz}{2}=\dfrac{s}{2}(x+y+z)</math>. Therefore, <math>\dfrac{s}{2}(x+y+z)=\dfrac{s}{2}(a)</math>, so <math>x+y+z=a</math>, which means the sum of the altitudes from any point within the triangle is equal to the altitude from the vertex of a triangle. | ||
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==See also== | ==See also== | ||
[[Category:Theorems]] | [[Category:Theorems]] | ||
Revision as of 11:24, 30 May 2019
The Viviani's Theorem states that for an equilateral triangle, the sum of the altitudes from any point in the triangle is equal to the altitude from a vertex of the triangle to the other side.
Proof
Let 
be an equilateral triangle and 
be a point inside the triangle.
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.88, xmax = 30.12, ymin = -5.5, ymax = 11.; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((3.22,-0.78)--(14.52,-0.7)--(8.800717967697247,9.046087062764157)--cycle, zzttqq); draw((10.222698397686688,-0.3061486732450197)--(9.798444960927041,-0.30915223739907016)--(9.801448525081092,-0.7334056741587179)--(10.22570196184074,-0.7304021100046675)--cycle, qqwuqq); draw((6.871598765286669,4.790362710570685)--(7.08112432080719,5.15927874652442)--(6.712208284853454,5.368804302044939)--(6.502682729332935,4.9998882660912045)--cycle, qqwuqq); draw((11.476061820875268,3.6488294940564145)--(11.690789702114396,3.28291702225673)--(12.05670217391408,3.4976449034958583)--(11.841974292674951,3.8635573752955428)--cycle, qqwuqq); /* draw figures */ draw((3.22,-0.78)--(14.52,-0.7), zzttqq); draw((14.52,-0.7)--(8.800717967697247,9.046087062764157), zzttqq); draw((8.800717967697247,9.046087062764157)--(3.22,-0.78), zzttqq); draw((10.2,2.9)--(10.22570196184074,-0.7304021100046675)); draw((10.2,2.9)--(6.502682729332935,4.9998882660912045)); draw((10.2,2.9)--(11.841974292674951,3.8635573752955428)); label("$x+y+z = a$",(7.5043433217971725,-1.565215213000298),SE*labelscalefactor); /* dots and labels */ dot((3.22,-0.78),dotstyle); label("$A$", (3.3,-0.58), NW * labelscalefactor); dot((14.52,-0.7),dotstyle); label("$B$", (14.6,-0.5), NE * labelscalefactor); dot((8.800717967697247,9.046087062764157),dotstyle); label("$C$", (8.88,9.24), N * labelscalefactor); dot((10.2,2.9),dotstyle); label("$P$", (10.28,3.1), NE * labelscalefactor); dot((6.502682729332935,4.9998882660912045),linewidth(3.pt) + dotstyle); label("$B'$", (6.58,5.12), NW * labelscalefactor); dot((11.841974292674951,3.8635573752955428),linewidth(3.pt) + dotstyle); label("$A'$", (11.92,3.98), NE * labelscalefactor); dot((10.22570196184074,-0.7304021100046675),linewidth(3.pt) + dotstyle); label("$C'$", (10.3,-0.62), NE * labelscalefactor); label("$x$", (9.88,1.1), NE * labelscalefactor); label("$z$", (8.5,4.24), NE * labelscalefactor); label("$y$", (11.18,3.12), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]](http://latex.artofproblemsolving.com/9/4/b/94bcbf3a28ea29197755181effb003894f14fe25.png)
We label the altitudes from to each of sides 
, 
and 
, 
and 
respectively. Since is equilateral, we can say that 
. Therefore, ![$[ABP]=\dfrac{sx}{2}$](http://latex.artofproblemsolving.com/3/e/1/3e1965aab6649a16f2534cfd3003f135a438096b.png)
, ![$[BCP]=\dfrac{sy}{2}$](http://latex.artofproblemsolving.com/3/8/0/380dcf10a2b74e7a65a5db42157b3efb498de67f.png)
and ![$[ACP]=\dfrac{sz}{2}$](http://latex.artofproblemsolving.com/d/8/6/d8600fc6537941ed1426e92a870f878929cf3c8a.png)
. Since the area of a triangle is the product of its base and altitude, we also have ![$[ABC]=\dfrac{as}{2}$](http://latex.artofproblemsolving.com/6/b/7/6b7488e59253a02cc677381a389700bea38ed51f.png)
. However, the area of can also be expressed as ![$[ABC]=[ABP]+[BCP]+[ACP]=\dfrac{sx}{2}+\dfrac{sy}{2}+\dfrac{sz}{2}=\dfrac{s}{2}(x+y+z)$](http://latex.artofproblemsolving.com/5/9/4/594cc7628b983dd48312f008f71de96e83c80c16.png)
. Therefore, 
, so 
, which means the sum of the altitudes from any point within the triangle is equal to the altitude from the vertex of a triangle.
