In triangle ABC D is a point on AC such that the in-circle of triangle ABD and BCD have equal radii. Find AD:DC.

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

In triangle ABC, it is known that AB = 12,BC = 13 and AC = 15. D is a point on AC such that the in-circle of triangle ABD and BCD have equal radii. Find AD:DC.
What i tried that as R is equal and R^2 = (S-A)(S-C)(S-B)/S so in this question
(let S1 be the semi-perimeter of triangle ABD and S2 of BCD)
(S1 - 12)(S1- AD)(S1-BD)/S1 = (S2-13)(S2-BD)(S2-DC)/S2

and then compare the terms. On comparing i get AD - DC = 1; and as AD + DC = 15 so
by this method i get AD:DC = 8:7 so is this method correct. Means can i compare LHS and RHS in this question the way i did. Or is there some other method to do this question.

geometry

share|cite|improve this question

asked Aug 6 at 13:23

Smit Patel

444

add a comment | 

up vote
1
down vote

favorite

In triangle ABC, it is known that AB = 12,BC = 13 and AC = 15. D is a point on AC such that the in-circle of triangle ABD and BCD have equal radii. Find AD:DC.
What i tried that as R is equal and R^2 = (S-A)(S-C)(S-B)/S so in this question
(let S1 be the semi-perimeter of triangle ABD and S2 of BCD)
(S1 - 12)(S1- AD)(S1-BD)/S1 = (S2-13)(S2-BD)(S2-DC)/S2

and then compare the terms. On comparing i get AD - DC = 1; and as AD + DC = 15 so
by this method i get AD:DC = 8:7 so is this method correct. Means can i compare LHS and RHS in this question the way i did. Or is there some other method to do this question.

geometry

share|cite|improve this question

asked Aug 6 at 13:23

Smit Patel

444

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

In triangle ABC, it is known that AB = 12,BC = 13 and AC = 15. D is a point on AC such that the in-circle of triangle ABD and BCD have equal radii. Find AD:DC.
What i tried that as R is equal and R^2 = (S-A)(S-C)(S-B)/S so in this question
(let S1 be the semi-perimeter of triangle ABD and S2 of BCD)
(S1 - 12)(S1- AD)(S1-BD)/S1 = (S2-13)(S2-BD)(S2-DC)/S2

and then compare the terms. On comparing i get AD - DC = 1; and as AD + DC = 15 so
by this method i get AD:DC = 8:7 so is this method correct. Means can i compare LHS and RHS in this question the way i did. Or is there some other method to do this question.

geometry

share|cite|improve this question

asked Aug 6 at 13:23

Smit Patel

444

In triangle ABC, it is known that AB = 12,BC = 13 and AC = 15. D is a point on AC such that the in-circle of triangle ABD and BCD have equal radii. Find AD:DC.
What i tried that as R is equal and R^2 = (S-A)(S-C)(S-B)/S so in this question
(let S1 be the semi-perimeter of triangle ABD and S2 of BCD)
(S1 - 12)(S1- AD)(S1-BD)/S1 = (S2-13)(S2-BD)(S2-DC)/S2

and then compare the terms. On comparing i get AD - DC = 1; and as AD + DC = 15 so
by this method i get AD:DC = 8:7 so is this method correct. Means can i compare LHS and RHS in this question the way i did. Or is there some other method to do this question.

geometry

share|cite|improve this question

asked Aug 6 at 13:23

Smit Patel

444

share|cite|improve this question

share|cite|improve this question

asked Aug 6 at 13:23

Smit Patel

444

asked Aug 6 at 13:23

Smit Patel

444

asked Aug 6 at 13:23

Smit Patel

444

444

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote

Solution

Lemma

Let $M$ be a point on the side $BC$ of $triangle ABC$ such that the
radiuses of the incircles of $triangle ABM$ and $triangle ACM$ are
equal. Then $$AM=fracsqrt(b+c)^2-a^22,$$ where $a,b,c~~$ are the
lengthes of $BC, CA, AB$ respectively.

As for the present problem, we obtain $$BD=fracsqrt(12+13)^2-15^22=10.$$

Notice that $$fracS_ABDS_BCD=fracAB+BD+ADBC+BD+DC=fracADDC.$$

Namely, $$frac22+AD23+DC=fracADDC.$$

But $$AD+DC=15.$$

Therefore, $$AD=frac223,~~~DC=frac233.$$

It follows that $$fracADDC=frac2223.$$

share|cite|improve this answer

edited Aug 6 at 15:26

answered Aug 6 at 15:14

mengdie1982

2,972216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you found out that S (of ABD triange )/S(of BCD triangle)=(AB+BD+AD)/(BC+BD+DC)=AD/DC.
  – Smit Patel
  Aug 6 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $S=pr$, where $p$ is the semi-perimeter. but $r$ is same....
  – mengdie1982
  Aug 6 at 17:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is S? And if S is area then how area of triangle ABD/BDC = AD/DC
  – Smit Patel
  Aug 6 at 17:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  don't you see the two triangle have the same altitude？
  – mengdie1982
  Aug 6 at 17:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  very sorry i got it
  – Smit Patel
  Aug 7 at 7:55
  

  
  
  
  

 | 
show 3 more comments

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873863%2fin-triangle-abc-d-is-a-point-on-ac-such-that-the-in-circle-of-triangle-abd-and-b%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Solution

Lemma

Let $M$ be a point on the side $BC$ of $triangle ABC$ such that the
radiuses of the incircles of $triangle ABM$ and $triangle ACM$ are
equal. Then $$AM=fracsqrt(b+c)^2-a^22,$$ where $a,b,c~~$ are the
lengthes of $BC, CA, AB$ respectively.

As for the present problem, we obtain $$BD=fracsqrt(12+13)^2-15^22=10.$$

Notice that $$fracS_ABDS_BCD=fracAB+BD+ADBC+BD+DC=fracADDC.$$

Namely, $$frac22+AD23+DC=fracADDC.$$

But $$AD+DC=15.$$

Therefore, $$AD=frac223,~~~DC=frac233.$$

It follows that $$fracADDC=frac2223.$$

share|cite|improve this answer

edited Aug 6 at 15:26

answered Aug 6 at 15:14

mengdie1982

2,972216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you found out that S (of ABD triange )/S(of BCD triangle)=(AB+BD+AD)/(BC+BD+DC)=AD/DC.
  – Smit Patel
  Aug 6 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $S=pr$, where $p$ is the semi-perimeter. but $r$ is same....
  – mengdie1982
  Aug 6 at 17:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is S? And if S is area then how area of triangle ABD/BDC = AD/DC
  – Smit Patel
  Aug 6 at 17:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  don't you see the two triangle have the same altitude？
  – mengdie1982
  Aug 6 at 17:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  very sorry i got it
  – Smit Patel
  Aug 7 at 7:55
  

  
  
  
  

 | 
show 3 more comments

up vote
1
down vote

Solution

Lemma

Let $M$ be a point on the side $BC$ of $triangle ABC$ such that the
radiuses of the incircles of $triangle ABM$ and $triangle ACM$ are
equal. Then $$AM=fracsqrt(b+c)^2-a^22,$$ where $a,b,c~~$ are the
lengthes of $BC, CA, AB$ respectively.

As for the present problem, we obtain $$BD=fracsqrt(12+13)^2-15^22=10.$$

Notice that $$fracS_ABDS_BCD=fracAB+BD+ADBC+BD+DC=fracADDC.$$

Namely, $$frac22+AD23+DC=fracADDC.$$

But $$AD+DC=15.$$

Therefore, $$AD=frac223,~~~DC=frac233.$$

It follows that $$fracADDC=frac2223.$$

share|cite|improve this answer

edited Aug 6 at 15:26

answered Aug 6 at 15:14

mengdie1982

2,972216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you found out that S (of ABD triange )/S(of BCD triangle)=(AB+BD+AD)/(BC+BD+DC)=AD/DC.
  – Smit Patel
  Aug 6 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $S=pr$, where $p$ is the semi-perimeter. but $r$ is same....
  – mengdie1982
  Aug 6 at 17:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is S? And if S is area then how area of triangle ABD/BDC = AD/DC
  – Smit Patel
  Aug 6 at 17:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  don't you see the two triangle have the same altitude？
  – mengdie1982
  Aug 6 at 17:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  very sorry i got it
  – Smit Patel
  Aug 7 at 7:55
  

  
  
  
  

 | 
show 3 more comments

up vote
1
down vote

up vote
1
down vote

Solution

Lemma

Let $M$ be a point on the side $BC$ of $triangle ABC$ such that the
radiuses of the incircles of $triangle ABM$ and $triangle ACM$ are
equal. Then $$AM=fracsqrt(b+c)^2-a^22,$$ where $a,b,c~~$ are the
lengthes of $BC, CA, AB$ respectively.

As for the present problem, we obtain $$BD=fracsqrt(12+13)^2-15^22=10.$$

Notice that $$fracS_ABDS_BCD=fracAB+BD+ADBC+BD+DC=fracADDC.$$

Namely, $$frac22+AD23+DC=fracADDC.$$

But $$AD+DC=15.$$

Therefore, $$AD=frac223,~~~DC=frac233.$$

It follows that $$fracADDC=frac2223.$$

share|cite|improve this answer

edited Aug 6 at 15:26

answered Aug 6 at 15:14

mengdie1982

2,972216

Solution

Lemma

Let $M$ be a point on the side $BC$ of $triangle ABC$ such that the
radiuses of the incircles of $triangle ABM$ and $triangle ACM$ are
equal. Then $$AM=fracsqrt(b+c)^2-a^22,$$ where $a,b,c~~$ are the
lengthes of $BC, CA, AB$ respectively.

As for the present problem, we obtain $$BD=fracsqrt(12+13)^2-15^22=10.$$

Notice that $$fracS_ABDS_BCD=fracAB+BD+ADBC+BD+DC=fracADDC.$$

Namely, $$frac22+AD23+DC=fracADDC.$$

But $$AD+DC=15.$$

Therefore, $$AD=frac223,~~~DC=frac233.$$

It follows that $$fracADDC=frac2223.$$

share|cite|improve this answer

edited Aug 6 at 15:26

answered Aug 6 at 15:14

mengdie1982

2,972216

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 15:26

edited Aug 6 at 15:26

edited Aug 6 at 15:26

answered Aug 6 at 15:14

mengdie1982

2,972216

answered Aug 6 at 15:14

mengdie1982

2,972216

answered Aug 6 at 15:14

mengdie1982

2,972216

2,972216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you found out that S (of ABD triange )/S(of BCD triangle)=(AB+BD+AD)/(BC+BD+DC)=AD/DC.
  – Smit Patel
  Aug 6 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $S=pr$, where $p$ is the semi-perimeter. but $r$ is same....
  – mengdie1982
  Aug 6 at 17:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is S? And if S is area then how area of triangle ABD/BDC = AD/DC
  – Smit Patel
  Aug 6 at 17:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  don't you see the two triangle have the same altitude？
  – mengdie1982
  Aug 6 at 17:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  very sorry i got it
  – Smit Patel
  Aug 7 at 7:55
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you found out that S (of ABD triange )/S(of BCD triangle)=(AB+BD+AD)/(BC+BD+DC)=AD/DC.
  – Smit Patel
  Aug 6 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $S=pr$, where $p$ is the semi-perimeter. but $r$ is same....
  – mengdie1982
  Aug 6 at 17:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is S? And if S is area then how area of triangle ABD/BDC = AD/DC
  – Smit Patel
  Aug 6 at 17:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  don't you see the two triangle have the same altitude？
  – mengdie1982
  Aug 6 at 17:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  very sorry i got it
  – Smit Patel
  Aug 7 at 7:55
  

  
  
  
  

How did you found out that S (of ABD triange )/S(of BCD triangle)=(AB+BD+AD)/(BC+BD+DC)=AD/DC.
– Smit Patel
Aug 6 at 16:49

How did you found out that S (of ABD triange )/S(of BCD triangle)=(AB+BD+AD)/(BC+BD+DC)=AD/DC.
– Smit Patel
Aug 6 at 16:49

$S=pr$, where $p$ is the semi-perimeter. but $r$ is same....
– mengdie1982
Aug 6 at 17:01

$S=pr$, where $p$ is the semi-perimeter. but $r$ is same....
– mengdie1982
Aug 6 at 17:01

What is S? And if S is area then how area of triangle ABD/BDC = AD/DC
– Smit Patel
Aug 6 at 17:16

What is S? And if S is area then how area of triangle ABD/BDC = AD/DC
– Smit Patel
Aug 6 at 17:16

don't you see the two triangle have the same altitude？
– mengdie1982
Aug 6 at 17:34

don't you see the two triangle have the same altitude？
– mengdie1982
Aug 6 at 17:34

very sorry i got it
– Smit Patel
Aug 7 at 7:55

very sorry i got it
– Smit Patel
Aug 7 at 7:55

 | 
show 3 more comments

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873863%2fin-triangle-abc-d-is-a-point-on-ac-such-that-the-in-circle-of-triangle-abd-and-b%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email