Understanding Ceva's Theorem Announcing the arrival of Valued Associate #679: Cesar Manara ...
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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 $frac{A_{triangle PXB}}{A_{triangle PXC}}=frac{BX}{CX}=frac{A_{triangle BXA}}{A_{triangle CXA}}$.
I would like clarification in understanding the following step which states:
$frac{A_{triangle APB}}{A_{triangle APC}}=frac{A_{triangle AXB} - A_{triangle PXB}}{A_{triangle AXC}-A_{triangle PXC}}=frac{BX}{CX}$
How does the subtraction of the two areas make it so that the new triangles are still proportional to $frac{BX}{CX}$? (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 $frac{A_{triangle PXB}}{A_{triangle PXC}}=frac{BX}{CX}=frac{A_{triangle BXA}}{A_{triangle CXA}}$.
I would like clarification in understanding the following step which states:
$frac{A_{triangle APB}}{A_{triangle APC}}=frac{A_{triangle AXB} - A_{triangle PXB}}{A_{triangle AXC}-A_{triangle PXC}}=frac{BX}{CX}$
How does the subtraction of the two areas make it so that the new triangles are still proportional to $frac{BX}{CX}$? (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 $frac{A_{triangle PXB}}{A_{triangle PXC}}=frac{BX}{CX}=frac{A_{triangle BXA}}{A_{triangle CXA}}$.
I would like clarification in understanding the following step which states:
$frac{A_{triangle APB}}{A_{triangle APC}}=frac{A_{triangle AXB} - A_{triangle PXB}}{A_{triangle AXC}-A_{triangle PXC}}=frac{BX}{CX}$
How does the subtraction of the two areas make it so that the new triangles are still proportional to $frac{BX}{CX}$? (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 $frac{A_{triangle PXB}}{A_{triangle PXC}}=frac{BX}{CX}=frac{A_{triangle BXA}}{A_{triangle CXA}}$.
I would like clarification in understanding the following step which states:
$frac{A_{triangle APB}}{A_{triangle APC}}=frac{A_{triangle AXB} - A_{triangle PXB}}{A_{triangle AXC}-A_{triangle PXC}}=frac{BX}{CX}$
How does the subtraction of the two areas make it so that the new triangles are still proportional to $frac{BX}{CX}$? (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
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oldest
votes
$begingroup$
$A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$
$A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$
Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$
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
1
active
oldest
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$begingroup$
$A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$
$A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$
Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
$endgroup$
add a comment |
$begingroup$
$A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$
$A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$
Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
$endgroup$
add a comment |
$begingroup$
$A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$
$A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$
Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$
answered 2 hours ago
YuiTo ChengYuiTo Cheng
2,48341037
$endgroup$
$A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$
$A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$
Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$
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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