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Understanding Ceva's Theorem

6

1

$begingroup$

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

share|cite|improve this question

edited 4 hours ago

YuiTo Cheng

2,50341037

asked 4 hours ago

dragonkingdragonking

434

$endgroup$

add a comment | 

6

1

$begingroup$

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

share|cite|improve this question

edited 4 hours ago

YuiTo Cheng

2,50341037

asked 4 hours ago

dragonkingdragonking

434

$endgroup$

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

share|cite|improve this question

edited 4 hours ago

YuiTo Cheng

2,50341037

asked 4 hours ago

dragonkingdragonking

434

$endgroup$

share|cite|improve this question

edited 4 hours ago

YuiTo Cheng

2,50341037

asked 4 hours ago

dragonkingdragonking

434

share|cite|improve this question

edited 4 hours ago

YuiTo Cheng

2,50341037

asked 4 hours ago

dragonkingdragonking

434

edited 4 hours ago

YuiTo Cheng

2,50341037

edited 4 hours ago

YuiTo Cheng

2,50341037

asked 4 hours ago

dragonkingdragonking

434

asked 4 hours ago

dragonkingdragonking

434

1 Answer
1

active

oldest

votes

5

$begingroup$

$A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$

$A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$

Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$

share|cite|improve this answer

answered 4 hours ago

YuiTo ChengYuiTo Cheng

2,50341037

$endgroup$

add a comment | 

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5

$begingroup$

$A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$

$A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$

Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$

share|cite|improve this answer

answered 4 hours ago

YuiTo ChengYuiTo Cheng

2,50341037

$endgroup$

add a comment | 

5

$begingroup$

$A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$

$A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$

Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$

share|cite|improve this answer

answered 4 hours ago

YuiTo ChengYuiTo Cheng

2,50341037

$endgroup$

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 $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$

share|cite|improve this answer

answered 4 hours ago

YuiTo ChengYuiTo Cheng

2,50341037

$endgroup$

share|cite|improve this answer

answered 4 hours ago

YuiTo ChengYuiTo Cheng

2,50341037

answered 4 hours ago

YuiTo ChengYuiTo Cheng

2,50341037

answered 4 hours ago

YuiTo ChengYuiTo Cheng

2,50341037