Proving a line perpendicular to another in a circle

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I got a question at an exam recently, and was unable to solve it.

Let $Gamma_1$ be a circle with a chord $CD$ and a diameter $ABperp CD$ at $N$, with $ANgeq BN$. A circle $Gamma_2$ is drawn with centre $C$ and radius $CN$. It intersects $Gamma_1$ at $P,Q$. $PQ$ intersects $CD$ at $M$ and $AC$ at $K$. $NK$ extended meets $Gamma_2$ at $L$. Prove that $ALperp PQ$.

Here's the diagram:

I tried various approaches, like proving $AL$ tangent to $Gamma_2$, Proving $OC||AL$, proving $QP||BF$, proving $QB=PF$, proving $CNAL$ cyclic, all to no avail.

Please help.

geometry contest-math

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asked Aug 3 at 16:08

MalayTheDynamo

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I got a question at an exam recently, and was unable to solve it.

Let $Gamma_1$ be a circle with a chord $CD$ and a diameter $ABperp CD$ at $N$, with $ANgeq BN$. A circle $Gamma_2$ is drawn with centre $C$ and radius $CN$. It intersects $Gamma_1$ at $P,Q$. $PQ$ intersects $CD$ at $M$ and $AC$ at $K$. $NK$ extended meets $Gamma_2$ at $L$. Prove that $ALperp PQ$.

Here's the diagram:

I tried various approaches, like proving $AL$ tangent to $Gamma_2$, Proving $OC||AL$, proving $QP||BF$, proving $QB=PF$, proving $CNAL$ cyclic, all to no avail.

Please help.

geometry contest-math

share|cite|improve this question

asked Aug 3 at 16:08

MalayTheDynamo

2,011732

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

I got a question at an exam recently, and was unable to solve it.

Let $Gamma_1$ be a circle with a chord $CD$ and a diameter $ABperp CD$ at $N$, with $ANgeq BN$. A circle $Gamma_2$ is drawn with centre $C$ and radius $CN$. It intersects $Gamma_1$ at $P,Q$. $PQ$ intersects $CD$ at $M$ and $AC$ at $K$. $NK$ extended meets $Gamma_2$ at $L$. Prove that $ALperp PQ$.

Here's the diagram:

I tried various approaches, like proving $AL$ tangent to $Gamma_2$, Proving $OC||AL$, proving $QP||BF$, proving $QB=PF$, proving $CNAL$ cyclic, all to no avail.

Please help.

geometry contest-math

share|cite|improve this question

asked Aug 3 at 16:08

MalayTheDynamo

2,011732

I got a question at an exam recently, and was unable to solve it.

Let $Gamma_1$ be a circle with a chord $CD$ and a diameter $ABperp CD$ at $N$, with $ANgeq BN$. A circle $Gamma_2$ is drawn with centre $C$ and radius $CN$. It intersects $Gamma_1$ at $P,Q$. $PQ$ intersects $CD$ at $M$ and $AC$ at $K$. $NK$ extended meets $Gamma_2$ at $L$. Prove that $ALperp PQ$.

Here's the diagram:

I tried various approaches, like proving $AL$ tangent to $Gamma_2$, Proving $OC||AL$, proving $QP||BF$, proving $QB=PF$, proving $CNAL$ cyclic, all to no avail.

Please help.

geometry contest-math

share|cite|improve this question

asked Aug 3 at 16:08

MalayTheDynamo

2,011732

share|cite|improve this question

share|cite|improve this question

asked Aug 3 at 16:08

MalayTheDynamo

2,011732

asked Aug 3 at 16:08

MalayTheDynamo

2,011732

asked Aug 3 at 16:08

MalayTheDynamo

2,011732

2,011732

add a comment | 

add a comment | 

2 Answers
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Proof

Let $CD$ intersect $Gamma_2$ at a second point $S$. Notice that $PQ$ is the radical axis of $Gamma_1$ and $Gamma_2$. Thus $$(overlineCN+overlineCM)cdot(overlineCN-overlineCM) =overlineSMcdot overlineMN=overlineCMcdot overlineMD=(overlineCN-overlineMN)cdot(overlineCN+overlineMN), $$which implies $$overlineCM=overlineMN.tag1$$
Morover, since$$overlineCKcdotoverlineKA=overlineNKcdotoverlineKL,$$
then $A,N,C,L$ are concyclic. Therefore $$angle ALC=angle BNC=90^o,$$ which shows that $AL$ is tangent to $Gamma_2$. By the symmetry of the tangents $AL,AN$, we may obtain $$overlineLK=overlineKN.tag2$$
By $(1),(2)$,we have $$PQ // LCperp AL.$$

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edited Aug 3 at 17:45

answered Aug 3 at 17:19

mengdie1982

2,815216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be $CKcdot KA=NKcdot KL$?
  – MalayTheDynamo
  Aug 3 at 17:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MalayTheDynamo yes. Corrected.
  – mengdie1982
  Aug 3 at 17:44
  

  
  
  
  

add a comment | 

up vote
0
down vote

Diagram

The diagram shows, $AG parallel DF$ so $ChatFL = EhatAG = delta$ and $EhatMF = EhatGA = gamma$. We can note that $Cl perp AL$ and $Delta CLF$ and $CL perp AL$ then $gamma + delta = 90°$

share|cite|improve this answer

edited Aug 3 at 18:24

answered Aug 3 at 18:12

GinoCHJ

12

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please elaborate.
  – MalayTheDynamo
  Aug 3 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I changed. It is more clear, I think
  – GinoCHJ
  Aug 3 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How is $CLperp AL$?
  – MalayTheDynamo
  Aug 4 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $Delta ANC$ is similar than $Delta NKA$ ($KhatNA = 90° - beta$) than $NK perp AK$, then $NK = KL$ hence $CL perp AL$
  – GinoCHJ
  2 days ago
  

  
  
  
  

add a comment | 

Your Answer

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

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up vote
1
down vote

Proof

Let $CD$ intersect $Gamma_2$ at a second point $S$. Notice that $PQ$ is the radical axis of $Gamma_1$ and $Gamma_2$. Thus $$(overlineCN+overlineCM)cdot(overlineCN-overlineCM) =overlineSMcdot overlineMN=overlineCMcdot overlineMD=(overlineCN-overlineMN)cdot(overlineCN+overlineMN), $$which implies $$overlineCM=overlineMN.tag1$$
Morover, since$$overlineCKcdotoverlineKA=overlineNKcdotoverlineKL,$$
then $A,N,C,L$ are concyclic. Therefore $$angle ALC=angle BNC=90^o,$$ which shows that $AL$ is tangent to $Gamma_2$. By the symmetry of the tangents $AL,AN$, we may obtain $$overlineLK=overlineKN.tag2$$
By $(1),(2)$,we have $$PQ // LCperp AL.$$

share|cite|improve this answer

edited Aug 3 at 17:45

answered Aug 3 at 17:19

mengdie1982

2,815216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be $CKcdot KA=NKcdot KL$?
  – MalayTheDynamo
  Aug 3 at 17:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MalayTheDynamo yes. Corrected.
  – mengdie1982
  Aug 3 at 17:44
  

  
  
  
  

add a comment | 

up vote
1
down vote

Proof

Let $CD$ intersect $Gamma_2$ at a second point $S$. Notice that $PQ$ is the radical axis of $Gamma_1$ and $Gamma_2$. Thus $$(overlineCN+overlineCM)cdot(overlineCN-overlineCM) =overlineSMcdot overlineMN=overlineCMcdot overlineMD=(overlineCN-overlineMN)cdot(overlineCN+overlineMN), $$which implies $$overlineCM=overlineMN.tag1$$
Morover, since$$overlineCKcdotoverlineKA=overlineNKcdotoverlineKL,$$
then $A,N,C,L$ are concyclic. Therefore $$angle ALC=angle BNC=90^o,$$ which shows that $AL$ is tangent to $Gamma_2$. By the symmetry of the tangents $AL,AN$, we may obtain $$overlineLK=overlineKN.tag2$$
By $(1),(2)$,we have $$PQ // LCperp AL.$$

share|cite|improve this answer

edited Aug 3 at 17:45

answered Aug 3 at 17:19

mengdie1982

2,815216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be $CKcdot KA=NKcdot KL$?
  – MalayTheDynamo
  Aug 3 at 17:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MalayTheDynamo yes. Corrected.
  – mengdie1982
  Aug 3 at 17:44
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Proof

Let $CD$ intersect $Gamma_2$ at a second point $S$. Notice that $PQ$ is the radical axis of $Gamma_1$ and $Gamma_2$. Thus $$(overlineCN+overlineCM)cdot(overlineCN-overlineCM) =overlineSMcdot overlineMN=overlineCMcdot overlineMD=(overlineCN-overlineMN)cdot(overlineCN+overlineMN), $$which implies $$overlineCM=overlineMN.tag1$$
Morover, since$$overlineCKcdotoverlineKA=overlineNKcdotoverlineKL,$$
then $A,N,C,L$ are concyclic. Therefore $$angle ALC=angle BNC=90^o,$$ which shows that $AL$ is tangent to $Gamma_2$. By the symmetry of the tangents $AL,AN$, we may obtain $$overlineLK=overlineKN.tag2$$
By $(1),(2)$,we have $$PQ // LCperp AL.$$

share|cite|improve this answer

edited Aug 3 at 17:45

answered Aug 3 at 17:19

mengdie1982

2,815216

Proof

Let $CD$ intersect $Gamma_2$ at a second point $S$. Notice that $PQ$ is the radical axis of $Gamma_1$ and $Gamma_2$. Thus $$(overlineCN+overlineCM)cdot(overlineCN-overlineCM) =overlineSMcdot overlineMN=overlineCMcdot overlineMD=(overlineCN-overlineMN)cdot(overlineCN+overlineMN), $$which implies $$overlineCM=overlineMN.tag1$$
Morover, since$$overlineCKcdotoverlineKA=overlineNKcdotoverlineKL,$$
then $A,N,C,L$ are concyclic. Therefore $$angle ALC=angle BNC=90^o,$$ which shows that $AL$ is tangent to $Gamma_2$. By the symmetry of the tangents $AL,AN$, we may obtain $$overlineLK=overlineKN.tag2$$
By $(1),(2)$,we have $$PQ // LCperp AL.$$

share|cite|improve this answer

edited Aug 3 at 17:45

answered Aug 3 at 17:19

mengdie1982

2,815216

share|cite|improve this answer

share|cite|improve this answer

edited Aug 3 at 17:45

edited Aug 3 at 17:45

edited Aug 3 at 17:45

answered Aug 3 at 17:19

mengdie1982

2,815216

answered Aug 3 at 17:19

mengdie1982

2,815216

answered Aug 3 at 17:19

mengdie1982

2,815216

2,815216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be $CKcdot KA=NKcdot KL$?
  – MalayTheDynamo
  Aug 3 at 17:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MalayTheDynamo yes. Corrected.
  – mengdie1982
  Aug 3 at 17:44
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be $CKcdot KA=NKcdot KL$?
  – MalayTheDynamo
  Aug 3 at 17:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MalayTheDynamo yes. Corrected.
  – mengdie1982
  Aug 3 at 17:44
  

  
  
  
  

Shouldn't it be $CKcdot KA=NKcdot KL$?
– MalayTheDynamo
Aug 3 at 17:41

Shouldn't it be $CKcdot KA=NKcdot KL$?
– MalayTheDynamo
Aug 3 at 17:41

@MalayTheDynamo yes. Corrected.
– mengdie1982
Aug 3 at 17:44

@MalayTheDynamo yes. Corrected.
– mengdie1982
Aug 3 at 17:44

add a comment | 

up vote
0
down vote

Diagram

The diagram shows, $AG parallel DF$ so $ChatFL = EhatAG = delta$ and $EhatMF = EhatGA = gamma$. We can note that $Cl perp AL$ and $Delta CLF$ and $CL perp AL$ then $gamma + delta = 90°$

share|cite|improve this answer

edited Aug 3 at 18:24

answered Aug 3 at 18:12

GinoCHJ

12

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please elaborate.
  – MalayTheDynamo
  Aug 3 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I changed. It is more clear, I think
  – GinoCHJ
  Aug 3 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How is $CLperp AL$?
  – MalayTheDynamo
  Aug 4 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $Delta ANC$ is similar than $Delta NKA$ ($KhatNA = 90° - beta$) than $NK perp AK$, then $NK = KL$ hence $CL perp AL$
  – GinoCHJ
  2 days ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Diagram

The diagram shows, $AG parallel DF$ so $ChatFL = EhatAG = delta$ and $EhatMF = EhatGA = gamma$. We can note that $Cl perp AL$ and $Delta CLF$ and $CL perp AL$ then $gamma + delta = 90°$

share|cite|improve this answer

edited Aug 3 at 18:24

answered Aug 3 at 18:12

GinoCHJ

12

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please elaborate.
  – MalayTheDynamo
  Aug 3 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I changed. It is more clear, I think
  – GinoCHJ
  Aug 3 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How is $CLperp AL$?
  – MalayTheDynamo
  Aug 4 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $Delta ANC$ is similar than $Delta NKA$ ($KhatNA = 90° - beta$) than $NK perp AK$, then $NK = KL$ hence $CL perp AL$
  – GinoCHJ
  2 days ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Diagram

The diagram shows, $AG parallel DF$ so $ChatFL = EhatAG = delta$ and $EhatMF = EhatGA = gamma$. We can note that $Cl perp AL$ and $Delta CLF$ and $CL perp AL$ then $gamma + delta = 90°$

share|cite|improve this answer

edited Aug 3 at 18:24

answered Aug 3 at 18:12

GinoCHJ

12

Diagram

The diagram shows, $AG parallel DF$ so $ChatFL = EhatAG = delta$ and $EhatMF = EhatGA = gamma$. We can note that $Cl perp AL$ and $Delta CLF$ and $CL perp AL$ then $gamma + delta = 90°$

share|cite|improve this answer

edited Aug 3 at 18:24

answered Aug 3 at 18:12

GinoCHJ

12

share|cite|improve this answer

share|cite|improve this answer

edited Aug 3 at 18:24

edited Aug 3 at 18:24

edited Aug 3 at 18:24

answered Aug 3 at 18:12

GinoCHJ

12

answered Aug 3 at 18:12

GinoCHJ

12

answered Aug 3 at 18:12

GinoCHJ

12

12

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please elaborate.
  – MalayTheDynamo
  Aug 3 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I changed. It is more clear, I think
  – GinoCHJ
  Aug 3 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How is $CLperp AL$?
  – MalayTheDynamo
  Aug 4 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $Delta ANC$ is similar than $Delta NKA$ ($KhatNA = 90° - beta$) than $NK perp AK$, then $NK = KL$ hence $CL perp AL$
  – GinoCHJ
  2 days ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please elaborate.
  – MalayTheDynamo
  Aug 3 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I changed. It is more clear, I think
  – GinoCHJ
  Aug 3 at 18:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How is $CLperp AL$?
  – MalayTheDynamo
  Aug 4 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $Delta ANC$ is similar than $Delta NKA$ ($KhatNA = 90° - beta$) than $NK perp AK$, then $NK = KL$ hence $CL perp AL$
  – GinoCHJ
  2 days ago
  

  
  
  
  

Please elaborate.
– MalayTheDynamo
Aug 3 at 18:15

Please elaborate.
– MalayTheDynamo
Aug 3 at 18:15

I changed. It is more clear, I think
– GinoCHJ
Aug 3 at 18:25

I changed. It is more clear, I think
– GinoCHJ
Aug 3 at 18:25

How is $CLperp AL$?
– MalayTheDynamo
Aug 4 at 4:58

How is $CLperp AL$?
– MalayTheDynamo
Aug 4 at 4:58

$Delta ANC$ is similar than $Delta NKA$ ($KhatNA = 90° - beta$) than $NK perp AK$, then $NK = KL$ hence $CL perp AL$
– GinoCHJ
2 days ago

$Delta ANC$ is similar than $Delta NKA$ ($KhatNA = 90° - beta$) than $NK perp AK$, then $NK = KL$ hence $CL perp AL$
– GinoCHJ
2 days ago

add a comment | 

 

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