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Understanding Ceva's Theorem
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Understanding Ceva's Theorem
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Ratio of areas of similar triangles given SSHow to calculate Fermat point in a triangle most efficiently?Prove this is a rectangleThe formula for the area of two triangles determined by the diagonals of a trapezoidUSAMO 2005, Problem3 (Triangle Geometry)- Is my solution correct?Corresponding side in similar trianglesMake isosceles triangle with matchsticksCutting a Triangle Through Its CentroidSimilar triangles and cross section integrals.Postulate or Theorem: The areas of similar plane figures are proportional their linear dimensions squared?
$begingroup$
In Ceva's Theorem, I understand that $fracA_triangle PXBA_triangle PXC=fracBXCX=fracA_triangle BXAA_triangle CXA$.
I would like clarification in understanding the following step which states:
$fracA_triangle APBA_triangle APC=fracA_triangle AXB - A_triangle PXBA_triangle AXC-A_triangle PXC=fracBXCX$
How does the subtraction of the two areas make it so that the new triangles are still proportional to $fracBXCX$? (even though they do not share those sides!)
geometry proof-verification triangles
edited 2 hours ago
YuiTo Cheng
2,48341037
asked 2 hours ago
dragonkingdragonking
384
$endgroup$
add a comment |
$begingroup$
In Ceva's Theorem, I understand that $fracA_triangle PXBA_triangle PXC=fracBXCX=fracA_triangle BXAA_triangle CXA$.
I would like clarification in understanding the following step which states:
$fracA_triangle APBA_triangle APC=fracA_triangle AXB - A_triangle PXBA_triangle AXC-A_triangle PXC=fracBXCX$
How does the subtraction of the two areas make it so that the new triangles are still proportional to $fracBXCX$? (even though they do not share those sides!)
geometry proof-verification triangles
edited 2 hours ago
YuiTo Cheng
2,48341037
asked 2 hours ago
dragonkingdragonking
384
$endgroup$
add a comment |
$begingroup$
In Ceva's Theorem, I understand that $fracA_triangle PXBA_triangle PXC=fracBXCX=fracA_triangle BXAA_triangle CXA$.
I would like clarification in understanding the following step which states:
$fracA_triangle APBA_triangle APC=fracA_triangle AXB - A_triangle PXBA_triangle AXC-A_triangle PXC=fracBXCX$
How does the subtraction of the two areas make it so that the new triangles are still proportional to $fracBXCX$? (even though they do not share those sides!)
geometry proof-verification triangles
edited 2 hours ago
YuiTo Cheng
2,48341037
asked 2 hours ago
dragonkingdragonking
384
$endgroup$
In Ceva's Theorem, I understand that $fracA_triangle PXBA_triangle PXC=fracBXCX=fracA_triangle BXAA_triangle CXA$.
I would like clarification in understanding the following step which states:
$fracA_triangle APBA_triangle APC=fracA_triangle AXB - A_triangle PXBA_triangle AXC-A_triangle PXC=fracBXCX$
How does the subtraction of the two areas make it so that the new triangles are still proportional to $fracBXCX$? (even though they do not share those sides!)
geometry proof-verification triangles
geometry proof-verification triangles
edited 2 hours ago
YuiTo Cheng
2,48341037
asked 2 hours ago
dragonkingdragonking
384
edited 2 hours ago
YuiTo Cheng
2,48341037
asked 2 hours ago
dragonkingdragonking
384
edited 2 hours ago
YuiTo Cheng
2,48341037
edited 2 hours ago
YuiTo Cheng
2,48341037
edited 2 hours ago
YuiTo Cheng
2,48341037
2,48341037
asked 2 hours ago
dragonkingdragonking
384
asked 2 hours ago
dragonkingdragonking
384
asked 2 hours ago
dragonkingdragonking
384
384
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$
$A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$
Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
$A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$
$A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$
Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
$endgroup$
add a comment |
$begingroup$
$A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$
$A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$
Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
$endgroup$
add a comment |
$begingroup$
$A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$
$A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$
Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
$endgroup$
$A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$
$A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$
Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
2,48341037
add a comment |
add a comment |
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