Cevians $AD$, $BE$, $CF$ are concurrent, as are cevians $DP$, $EQ$, $FR$; show that $AP$, $BQ$, $CR$ are...

3

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In $triangle ABC$, $D$, $E$, and $F$ are points on $BC$, $CA$, and $AB$, respectively, such that $AD$, $BE$, and $CF$ are concurrent lines. Points $P$, $Q$, and $R$ respectively on $EF$, $FD$, and $DE$ are such that $DP$, $EQ$, and $FR$ are concurrent. Prove that $AP$, $BQ$, and $CR$ are also concurrent.

I am really not getting how to proceed at all. I know I'm supposed to use Ceva's theorem but where do I apply it.

geometry euclidean-geometry triangle

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edited Jan 10 at 3:57

Matt Samuel

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asked Dec 14 '18 at 3:19

Shubhraneel PalShubhraneel Pal

46439

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closed as off-topic by José Carlos Santos, Lee David Chung Lin, A. Pongrácz, amWhy, RRL Jan 14 at 16:53

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3

$begingroup$

In $triangle ABC$, $D$, $E$, and $F$ are points on $BC$, $CA$, and $AB$, respectively, such that $AD$, $BE$, and $CF$ are concurrent lines. Points $P$, $Q$, and $R$ respectively on $EF$, $FD$, and $DE$ are such that $DP$, $EQ$, and $FR$ are concurrent. Prove that $AP$, $BQ$, and $CR$ are also concurrent.

I am really not getting how to proceed at all. I know I'm supposed to use Ceva's theorem but where do I apply it.

geometry euclidean-geometry triangle

share|cite|improve this question

edited Jan 10 at 3:57

Matt Samuel

38.4k63768

asked Dec 14 '18 at 3:19

Shubhraneel PalShubhraneel Pal

46439

$endgroup$

closed as off-topic by José Carlos Santos, Lee David Chung Lin, A. Pongrácz, amWhy, RRL Jan 14 at 16:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Lee David Chung Lin, A. Pongrácz, amWhy, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

3

3

3

3

$begingroup$

In $triangle ABC$, $D$, $E$, and $F$ are points on $BC$, $CA$, and $AB$, respectively, such that $AD$, $BE$, and $CF$ are concurrent lines. Points $P$, $Q$, and $R$ respectively on $EF$, $FD$, and $DE$ are such that $DP$, $EQ$, and $FR$ are concurrent. Prove that $AP$, $BQ$, and $CR$ are also concurrent.

I am really not getting how to proceed at all. I know I'm supposed to use Ceva's theorem but where do I apply it.

geometry euclidean-geometry triangle

share|cite|improve this question

edited Jan 10 at 3:57

Matt Samuel

38.4k63768

asked Dec 14 '18 at 3:19

Shubhraneel PalShubhraneel Pal

46439

$endgroup$

In $triangle ABC$, $D$, $E$, and $F$ are points on $BC$, $CA$, and $AB$, respectively, such that $AD$, $BE$, and $CF$ are concurrent lines. Points $P$, $Q$, and $R$ respectively on $EF$, $FD$, and $DE$ are such that $DP$, $EQ$, and $FR$ are concurrent. Prove that $AP$, $BQ$, and $CR$ are also concurrent.

I am really not getting how to proceed at all. I know I'm supposed to use Ceva's theorem but where do I apply it.

geometry euclidean-geometry triangle

geometry euclidean-geometry triangle

share|cite|improve this question

edited Jan 10 at 3:57

Matt Samuel

38.4k63768

asked Dec 14 '18 at 3:19

Shubhraneel PalShubhraneel Pal

46439

share|cite|improve this question

edited Jan 10 at 3:57

Matt Samuel

38.4k63768

asked Dec 14 '18 at 3:19

Shubhraneel PalShubhraneel Pal

46439

share|cite|improve this question

share|cite|improve this question

edited Jan 10 at 3:57

Matt Samuel

38.4k63768

edited Jan 10 at 3:57

Matt Samuel

38.4k63768

edited Jan 10 at 3:57

Matt Samuel

38.4k63768

38.4k63768

asked Dec 14 '18 at 3:19

Shubhraneel PalShubhraneel Pal

46439

asked Dec 14 '18 at 3:19

Shubhraneel PalShubhraneel Pal

46439

asked Dec 14 '18 at 3:19

Shubhraneel PalShubhraneel Pal

46439

46439

closed as off-topic by José Carlos Santos, Lee David Chung Lin, A. Pongrácz, amWhy, RRL Jan 14 at 16:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Lee David Chung Lin, A. Pongrácz, amWhy, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, Lee David Chung Lin, A. Pongrácz, amWhy, RRL Jan 14 at 16:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Lee David Chung Lin, A. Pongrácz, amWhy, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.

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1 Answer
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11

$begingroup$

Yes, Ceva's a good choice here. Let's recall another version of Ceva's theorem

In the above picture, the three lines are concurrent if and only if
$$frac{YI}{IZ}cdot frac{ZH}{XH} cdotfrac{XG}{YG} = 1.$$
Now, the ratios of the lengths are equal to the ratios of the areas of certain triangles, more specifically:
$$frac{YI}{IZ} = frac{triangle XYI}{triangle XZI},dots$$
here, we simply denote the triangle by its area. But
$$triangle XYI = frac12 XYcdot XI cdot sin (angle YXI), $$
and similar for $triangle XZI$. So
$$frac{YI}{IZ} = frac{XY}{XZ}cdot frac{sinangle XYI}{sin angle ZXI}.$$
Taking cyclic product, we have the sine version of Ceva's theorem: That the three lines are concurrent if and only if
the product of cyclic ratios of sines is equal to $1$.

Now that's what we will use in this problem:

So we want to show that
$$mathcal{F} = frac{sinangle BAP}{sinangle PAC}cdot frac{sinangle ACR}{sinangle BCR} cdot frac{sinangle{CBQ}}{sinangle ABQ} = 1.tag{1}$$
Now
$$frac{sinangle BAP}{sinangle PAC} cdot frac{AF}{AE} =frac{triangle AFP}{triangle AEP} = frac{FP}{PE},$$
and so on.
Taking the cyclic product of the above, we obtain
$$mathcal{F}cdot frac{AF}{AE}cdotfrac{BD}{BF}cdot frac{CE}{CD} = frac{FP}{PE}cdotfrac{DQ}{FQ}cdotfrac{ER}{DR}.$$
And (1) follows from classic Ceva's theorem with the given concurrences.

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answered Dec 14 '18 at 4:34

Quang HoangQuang Hoang

13.1k1233

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

11

$begingroup$

Yes, Ceva's a good choice here. Let's recall another version of Ceva's theorem

In the above picture, the three lines are concurrent if and only if
$$frac{YI}{IZ}cdot frac{ZH}{XH} cdotfrac{XG}{YG} = 1.$$
Now, the ratios of the lengths are equal to the ratios of the areas of certain triangles, more specifically:
$$frac{YI}{IZ} = frac{triangle XYI}{triangle XZI},dots$$
here, we simply denote the triangle by its area. But
$$triangle XYI = frac12 XYcdot XI cdot sin (angle YXI), $$
and similar for $triangle XZI$. So
$$frac{YI}{IZ} = frac{XY}{XZ}cdot frac{sinangle XYI}{sin angle ZXI}.$$
Taking cyclic product, we have the sine version of Ceva's theorem: That the three lines are concurrent if and only if
the product of cyclic ratios of sines is equal to $1$.

Now that's what we will use in this problem:

So we want to show that
$$mathcal{F} = frac{sinangle BAP}{sinangle PAC}cdot frac{sinangle ACR}{sinangle BCR} cdot frac{sinangle{CBQ}}{sinangle ABQ} = 1.tag{1}$$
Now
$$frac{sinangle BAP}{sinangle PAC} cdot frac{AF}{AE} =frac{triangle AFP}{triangle AEP} = frac{FP}{PE},$$
and so on.
Taking the cyclic product of the above, we obtain
$$mathcal{F}cdot frac{AF}{AE}cdotfrac{BD}{BF}cdot frac{CE}{CD} = frac{FP}{PE}cdotfrac{DQ}{FQ}cdotfrac{ER}{DR}.$$
And (1) follows from classic Ceva's theorem with the given concurrences.

share|cite|improve this answer

answered Dec 14 '18 at 4:34

Quang HoangQuang Hoang

13.1k1233

$endgroup$

add a comment | 

11

$begingroup$

Yes, Ceva's a good choice here. Let's recall another version of Ceva's theorem

In the above picture, the three lines are concurrent if and only if
$$frac{YI}{IZ}cdot frac{ZH}{XH} cdotfrac{XG}{YG} = 1.$$
Now, the ratios of the lengths are equal to the ratios of the areas of certain triangles, more specifically:
$$frac{YI}{IZ} = frac{triangle XYI}{triangle XZI},dots$$
here, we simply denote the triangle by its area. But
$$triangle XYI = frac12 XYcdot XI cdot sin (angle YXI), $$
and similar for $triangle XZI$. So
$$frac{YI}{IZ} = frac{XY}{XZ}cdot frac{sinangle XYI}{sin angle ZXI}.$$
Taking cyclic product, we have the sine version of Ceva's theorem: That the three lines are concurrent if and only if
the product of cyclic ratios of sines is equal to $1$.

Now that's what we will use in this problem:

So we want to show that
$$mathcal{F} = frac{sinangle BAP}{sinangle PAC}cdot frac{sinangle ACR}{sinangle BCR} cdot frac{sinangle{CBQ}}{sinangle ABQ} = 1.tag{1}$$
Now
$$frac{sinangle BAP}{sinangle PAC} cdot frac{AF}{AE} =frac{triangle AFP}{triangle AEP} = frac{FP}{PE},$$
and so on.
Taking the cyclic product of the above, we obtain
$$mathcal{F}cdot frac{AF}{AE}cdotfrac{BD}{BF}cdot frac{CE}{CD} = frac{FP}{PE}cdotfrac{DQ}{FQ}cdotfrac{ER}{DR}.$$
And (1) follows from classic Ceva's theorem with the given concurrences.

share|cite|improve this answer

answered Dec 14 '18 at 4:34

Quang HoangQuang Hoang

13.1k1233

$endgroup$

add a comment | 

11

11

11

$begingroup$

Yes, Ceva's a good choice here. Let's recall another version of Ceva's theorem

In the above picture, the three lines are concurrent if and only if
$$frac{YI}{IZ}cdot frac{ZH}{XH} cdotfrac{XG}{YG} = 1.$$
Now, the ratios of the lengths are equal to the ratios of the areas of certain triangles, more specifically:
$$frac{YI}{IZ} = frac{triangle XYI}{triangle XZI},dots$$
here, we simply denote the triangle by its area. But
$$triangle XYI = frac12 XYcdot XI cdot sin (angle YXI), $$
and similar for $triangle XZI$. So
$$frac{YI}{IZ} = frac{XY}{XZ}cdot frac{sinangle XYI}{sin angle ZXI}.$$
Taking cyclic product, we have the sine version of Ceva's theorem: That the three lines are concurrent if and only if
the product of cyclic ratios of sines is equal to $1$.

Now that's what we will use in this problem:

So we want to show that
$$mathcal{F} = frac{sinangle BAP}{sinangle PAC}cdot frac{sinangle ACR}{sinangle BCR} cdot frac{sinangle{CBQ}}{sinangle ABQ} = 1.tag{1}$$
Now
$$frac{sinangle BAP}{sinangle PAC} cdot frac{AF}{AE} =frac{triangle AFP}{triangle AEP} = frac{FP}{PE},$$
and so on.
Taking the cyclic product of the above, we obtain
$$mathcal{F}cdot frac{AF}{AE}cdotfrac{BD}{BF}cdot frac{CE}{CD} = frac{FP}{PE}cdotfrac{DQ}{FQ}cdotfrac{ER}{DR}.$$
And (1) follows from classic Ceva's theorem with the given concurrences.

share|cite|improve this answer

answered Dec 14 '18 at 4:34

Quang HoangQuang Hoang

13.1k1233

$endgroup$

Yes, Ceva's a good choice here. Let's recall another version of Ceva's theorem

In the above picture, the three lines are concurrent if and only if
$$frac{YI}{IZ}cdot frac{ZH}{XH} cdotfrac{XG}{YG} = 1.$$
Now, the ratios of the lengths are equal to the ratios of the areas of certain triangles, more specifically:
$$frac{YI}{IZ} = frac{triangle XYI}{triangle XZI},dots$$
here, we simply denote the triangle by its area. But
$$triangle XYI = frac12 XYcdot XI cdot sin (angle YXI), $$
and similar for $triangle XZI$. So
$$frac{YI}{IZ} = frac{XY}{XZ}cdot frac{sinangle XYI}{sin angle ZXI}.$$
Taking cyclic product, we have the sine version of Ceva's theorem: That the three lines are concurrent if and only if
the product of cyclic ratios of sines is equal to $1$.

Now that's what we will use in this problem:

So we want to show that
$$mathcal{F} = frac{sinangle BAP}{sinangle PAC}cdot frac{sinangle ACR}{sinangle BCR} cdot frac{sinangle{CBQ}}{sinangle ABQ} = 1.tag{1}$$
Now
$$frac{sinangle BAP}{sinangle PAC} cdot frac{AF}{AE} =frac{triangle AFP}{triangle AEP} = frac{FP}{PE},$$
and so on.
Taking the cyclic product of the above, we obtain
$$mathcal{F}cdot frac{AF}{AE}cdotfrac{BD}{BF}cdot frac{CE}{CD} = frac{FP}{PE}cdotfrac{DQ}{FQ}cdotfrac{ER}{DR}.$$
And (1) follows from classic Ceva's theorem with the given concurrences.

share|cite|improve this answer

answered Dec 14 '18 at 4:34

Quang HoangQuang Hoang

13.1k1233

share|cite|improve this answer

share|cite|improve this answer

answered Dec 14 '18 at 4:34

Quang HoangQuang Hoang

13.1k1233

answered Dec 14 '18 at 4:34

Quang HoangQuang Hoang

13.1k1233

answered Dec 14 '18 at 4:34

Quang HoangQuang Hoang

13.1k1233

13.1k1233

add a comment | 

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