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Understanding Ceva's Theorem

5

1

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

YuiTo Cheng

2,48341037

asked 3 hours ago

dragonkingdragonking

384

$endgroup$

add a comment | 

5

1

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

YuiTo Cheng

2,48341037

asked 3 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

share|cite|improve this question

edited 3 hours ago

YuiTo Cheng

2,48341037

asked 3 hours ago

dragonkingdragonking

384

$endgroup$

share|cite|improve this question

edited 3 hours ago

YuiTo Cheng

2,48341037

asked 3 hours ago

dragonkingdragonking

384

share|cite|improve this question

edited 3 hours ago

YuiTo Cheng

2,48341037

asked 3 hours ago

dragonkingdragonking

384

edited 3 hours ago

YuiTo Cheng

2,48341037

edited 3 hours ago

YuiTo Cheng

2,48341037

asked 3 hours ago

dragonkingdragonking

384

asked 3 hours ago

dragonkingdragonking

384

1 Answer
1

active

oldest

votes

3

$begingroup$

$A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$

$A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$

Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$

share|cite|improve this answer

answered 2 hours ago

YuiTo ChengYuiTo Cheng

2,48341037

$endgroup$

add a comment | 

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3

$begingroup$

$A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$

$A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$

Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$

share|cite|improve this answer

answered 2 hours ago

YuiTo ChengYuiTo Cheng

2,48341037

$endgroup$

add a comment | 

3

$begingroup$

$A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$

$A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$

Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$

share|cite|improve this answer

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 $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$

share|cite|improve this answer

answered 2 hours ago

YuiTo ChengYuiTo Cheng

2,48341037

$endgroup$

share|cite|improve this answer

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