Difference between revisions of "Viviani's theorem"
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| − | '''Viviani's Theorem''' states that for an equilateral triangle, the sum of the altitudes from | + | '''Viviani's Theorem''' states that for an equilateral triangle, the sum of the altitudes from an arbitrary point chosen within or on the triangle is equal to the height from a vertex of the triangle to the other side. |
== Proof == | == Proof == | ||
Latest revision as of 01:13, 5 February 2025
Viviani's Theorem states that for an equilateral triangle, the sum of the altitudes from an arbitrary point chosen within or on the triangle is equal to the height 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.
