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For a given triangle, prove that $DL=DM$

0

$begingroup$

In a triangle $ABC$, $D$ is midpoint of the side $BC$. Through the point $A$, $PQ$ is any straight line. The perpendiculars from the points $B$, $C$ and $D$ on $PQ$ are $BL$, $CM$ and $DN$ respectively. Prove that $DL = DM$

Can someone give me hint as how to use the information, '$D$ is the mid point of BC'? We can use the fact that $BL,DN,CM$ will be parallel to each other

geometry triangles

share|cite|improve this question

asked Mar 6 '16 at 8:55

AkiraAkira

3301211

$endgroup$

add a comment | 

0

$begingroup$

In a triangle $ABC$, $D$ is midpoint of the side $BC$. Through the point $A$, $PQ$ is any straight line. The perpendiculars from the points $B$, $C$ and $D$ on $PQ$ are $BL$, $CM$ and $DN$ respectively. Prove that $DL = DM$

Can someone give me hint as how to use the information, '$D$ is the mid point of BC'? We can use the fact that $BL,DN,CM$ will be parallel to each other

geometry triangles

share|cite|improve this question

asked Mar 6 '16 at 8:55

AkiraAkira

3301211

$endgroup$

add a comment | 

$begingroup$

In a triangle $ABC$, $D$ is midpoint of the side $BC$. Through the point $A$, $PQ$ is any straight line. The perpendiculars from the points $B$, $C$ and $D$ on $PQ$ are $BL$, $CM$ and $DN$ respectively. Prove that $DL = DM$

Can someone give me hint as how to use the information, '$D$ is the mid point of BC'? We can use the fact that $BL,DN,CM$ will be parallel to each other

geometry triangles

share|cite|improve this question

asked Mar 6 '16 at 8:55

AkiraAkira

3301211

$endgroup$

share|cite|improve this question

asked Mar 6 '16 at 8:55

AkiraAkira

3301211

share|cite|improve this question

asked Mar 6 '16 at 8:55

AkiraAkira

3301211

asked Mar 6 '16 at 8:55

AkiraAkira

3301211

asked Mar 6 '16 at 8:55

AkiraAkira

3301211

2 Answers
2

active

oldest

votes

1

$begingroup$

$triangle LDN = triangle MDN (LN=MN, ND - common, angle LND = angle MND= 90^circ ) Rightarrow DL=DM$

$DN - $ the middle line of the trapezoid $BLMC$ ($BL||DN||CM$ and $BD=CD$)

share|cite|improve this answer

edited Mar 6 '16 at 9:10

answered Mar 6 '16 at 9:06

Roman83Roman83

14.4k31956

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is $LN=MN$?
  $endgroup$
  – Akira
  Mar 6 '16 at 9:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $DN - $ the middle line of the trapezoid $BLMC$
  $endgroup$
  – Roman83
  Mar 6 '16 at 9:09
  

  
  
  
  

add a comment | 

0

$begingroup$

$BL||DN||CM$ and $BD=CD$

By intercept theorem $LN=LM$
Clearly $triangle LND$ and $triangle MND$ are congruent by S-A-S

So,$ DL=DM $ (Corresponding parts of congruent triangles)

share|cite|improve this answer

edited Apr 1 at 4:09

dantopa

6,69442245

answered Apr 1 at 3:45

Divya Prakash SinhaDivya Prakash Sinha

296

$endgroup$

add a comment | 

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1

$begingroup$

$triangle LDN = triangle MDN (LN=MN, ND - common, angle LND = angle MND= 90^circ ) Rightarrow DL=DM$

$DN - $ the middle line of the trapezoid $BLMC$ ($BL||DN||CM$ and $BD=CD$)

share|cite|improve this answer

edited Mar 6 '16 at 9:10

answered Mar 6 '16 at 9:06

Roman83Roman83

14.4k31956

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is $LN=MN$?
  $endgroup$
  – Akira
  Mar 6 '16 at 9:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $DN - $ the middle line of the trapezoid $BLMC$
  $endgroup$
  – Roman83
  Mar 6 '16 at 9:09
  

  
  
  
  

add a comment | 

1

$begingroup$

$triangle LDN = triangle MDN (LN=MN, ND - common, angle LND = angle MND= 90^circ ) Rightarrow DL=DM$

$DN - $ the middle line of the trapezoid $BLMC$ ($BL||DN||CM$ and $BD=CD$)

share|cite|improve this answer

edited Mar 6 '16 at 9:10

answered Mar 6 '16 at 9:06

Roman83Roman83

14.4k31956

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is $LN=MN$?
  $endgroup$
  – Akira
  Mar 6 '16 at 9:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $DN - $ the middle line of the trapezoid $BLMC$
  $endgroup$
  – Roman83
  Mar 6 '16 at 9:09
  

  
  
  
  

add a comment | 

$begingroup$

$triangle LDN = triangle MDN (LN=MN, ND - common, angle LND = angle MND= 90^circ ) Rightarrow DL=DM$

$DN - $ the middle line of the trapezoid $BLMC$ ($BL||DN||CM$ and $BD=CD$)

share|cite|improve this answer

edited Mar 6 '16 at 9:10

answered Mar 6 '16 at 9:06

Roman83Roman83

14.4k31956

$endgroup$

share|cite|improve this answer

edited Mar 6 '16 at 9:10

answered Mar 6 '16 at 9:06

Roman83Roman83

14.4k31956

answered Mar 6 '16 at 9:06

Roman83Roman83

14.4k31956

answered Mar 6 '16 at 9:06

Roman83Roman83

14.4k31956

0

$begingroup$

$BL||DN||CM$ and $BD=CD$

By intercept theorem $LN=LM$
Clearly $triangle LND$ and $triangle MND$ are congruent by S-A-S

So,$ DL=DM $ (Corresponding parts of congruent triangles)

share|cite|improve this answer

edited Apr 1 at 4:09

dantopa

6,69442245

answered Apr 1 at 3:45

Divya Prakash SinhaDivya Prakash Sinha

296

$endgroup$

add a comment | 

0

$begingroup$

$BL||DN||CM$ and $BD=CD$

By intercept theorem $LN=LM$
Clearly $triangle LND$ and $triangle MND$ are congruent by S-A-S

So,$ DL=DM $ (Corresponding parts of congruent triangles)

share|cite|improve this answer

edited Apr 1 at 4:09

dantopa

6,69442245

answered Apr 1 at 3:45

Divya Prakash SinhaDivya Prakash Sinha

296

$endgroup$

add a comment | 

$begingroup$

$BL||DN||CM$ and $BD=CD$

By intercept theorem $LN=LM$
Clearly $triangle LND$ and $triangle MND$ are congruent by S-A-S

So,$ DL=DM $ (Corresponding parts of congruent triangles)

share|cite|improve this answer

edited Apr 1 at 4:09

dantopa

6,69442245

answered Apr 1 at 3:45

Divya Prakash SinhaDivya Prakash Sinha

296

$endgroup$

share|cite|improve this answer

edited Apr 1 at 4:09

dantopa

6,69442245

answered Apr 1 at 3:45

Divya Prakash SinhaDivya Prakash Sinha

296

edited Apr 1 at 4:09

dantopa

6,69442245

edited Apr 1 at 4:09

dantopa

6,69442245

answered Apr 1 at 3:45

Divya Prakash SinhaDivya Prakash Sinha

296

answered Apr 1 at 3:45

Divya Prakash SinhaDivya Prakash Sinha

296