text: # Thread: Help with these two geometry!
  
  1. ## Help with these two geometry!
  
  Help with these two hard problems:
  
  Problem 12
  Let M be a point inside the angle A. From point M perpendiculars MP and MQ are dropped to the angle sides. From point A, perpendicular AK is dropped to segment PQ. Prove that angle PAK = angle MAQ.
  Problem 13
  Let ABC be a triangle and angle B = 60. Angle bisectors AD and CE meet at the point O. Prove that OD = OE.
  
  2. Hello xxravenxx
  Originally Posted by xxravenxx
  ... Problem 12
  Let M be a point inside the angle A. From point M perpendiculars MP and MQ are dropped to the angle sides. From point A, perpendicular AK is dropped to segment PQ. Prove that angle PAK = angle MAQ.
  $\angle APM = \angle AQM = 90^o$ (given)
  
  $\Rightarrow APMQ$ is a cyclic quadrilateral (opp angles supplementary)
  
  $\Rightarrow \angle APQ = \angle AMQ$ (angles in same segment)
  
  So in $\triangle$'s $APK, AMQ:$
  $\angle APK = \angle AMQ$ (proven)
  $\angle AKP = \angle AQM$ ( $=90^o$, given)
  $\Rightarrow \angle PAK = \angle MAQ$ (angle sum of triangle)
  
  3. Sorry but can you draw me a diagram of what this looks like? I'm a bit clueless.
  
  4. Hello xxravenxx
  Originally Posted by xxravenxx
  Sorry but can you draw me a diagram of what this looks like? I'm a bit clueless.
  See attachment.
  
  ... and here's the answer for number 13:
  Originally Posted by xxravenxx
  ...Problem 13
  Let ABC be a triangle and angle B = 60. Angle bisectors AD and CE meet at the point O. Prove that OD = OE.
  (I hope you can draw a diagram for this.)
  
  Let $\angle BAD = \angle DAC = x^o$ and $\angle ACE = \angle ECB = y^o$
  
  Then, $2x+2y+ 60=180$ (angle sum of $\triangle ABC$)
  
  $\Rightarrow x+y=60$
  
  Now $\angle DOC$ is an exterior angle of $\triangle OAC$ = sum of interior opposite angles $= x+y = 60^o$
  
  $\Rightarrow BEOD$ is a cyclic quadrilateral (exterior angle $\angle DOC$ = interior opposite angle $\angle B$)
  
  Now the angle bisectors of a triangle are concurrent $\Rightarrow BO$ is the bisector of $\angle ABC$
  
  $\Rightarrow \angle DBO = \angle EBO$
  
  $\Rightarrow \angle DEO = \angle EDO$ (angles in same segment)
  
  $\Rightarrow OE = OD$ (isosceles $\triangle OED$)
explanation: The document contains a thread from a forum where a user named xxravenxx posts two geometry problems and requests help. Another user provides detailed solutions to both problems, including a request for a diagram for the first problem and a step-by-step proof for both problems. The document includes a question and an answer for each problem.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Thread: Help with these two geometry!
  
  1. ## Help with these two geometry!
  
  Help with these two hard problems:
  
  Problem 12
  Let M be a point inside the angle A. From point M perpendiculars MP and MQ are dropped to the angle sides. From point A, perpendicular AK is dropped to segment PQ. Prove that angle PAK = angle MAQ.
  Problem 13
  Let ABC be a triangle and angle B = 60. Angle bisectors AD and CE meet at the point O. Prove that OD = OE.
  
  2. Hello xxravenxx
  Originally Posted by xxravenxx
  ... Problem 12
  Let M be a point inside the angle A. From point M perpendiculars MP and MQ are dropped to the angle sides. From point A, perpendicular AK is dropped to segment PQ. Prove that angle PAK = angle MAQ.
  $\angle APM = \angle AQM = 90^o$ (given)
  
  $\Rightarrow APMQ$ is a cyclic quadrilateral (opp angles supplementary)
  
  $\Rightarrow \angle APQ = \angle AMQ$ (angles in same segment)
  
  So in $\triangle$'s $APK, AMQ:$
  $\angle APK = \angle AMQ$ (proven)
  $\angle AKP = \angle AQM$ ( $=90^o$, given)
  $\Rightarrow \angle PAK = \angle MAQ$ (angle sum of triangle)
  
  3. Sorry but can you draw me a diagram of what this looks like? I'm a bit clueless.
  
  4. Hello xxravenxx
  Originally Posted by xxravenxx
  Sorry but can you draw me a diagram of what this looks like? I'm a bit clueless.
  See attachment.
  
  ... and here's the answer for number 13:
  Originally Posted by xxravenxx
  ...Problem 13
  Let ABC be a triangle and angle B = 60. Angle bisectors AD and CE meet at the point O. Prove that OD = OE.
  (I hope you can draw a diagram for this.)
  
  Let $\angle BAD = \angle DAC = x^o$ and $\angle ACE = \angle ECB = y^o$
  
  Then, $2x+2y+ 60=180$ (angle sum of $\triangle ABC$)
  
  $\Rightarrow x+y=60$
  
  Now $\angle DOC$ is an exterior angle of $\triangle OAC$ = sum of interior opposite angles $= x+y = 60^o$
  
  $\Rightarrow BEOD$ is a cyclic quadrilateral (exterior angle $\angle DOC$ = interior opposite angle $\angle B$)
  
  Now the angle bisectors of a triangle are concurrent $\Rightarrow BO$ is the bisector of $\angle ABC$
  
  $\Rightarrow \angle DBO = \angle EBO$
  
  $\Rightarrow \angle DEO = \angle EDO$ (angles in same segment)
  
  $\Rightarrow OE = OD$ (isosceles $\triangle OED$)
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a thread from a forum where a user named xxravenxx posts two geometry problems and requests help. Another user provides detailed solutions to both problems, including a request for a diagram for the first problem and a step-by-step proof for both problems. The document includes a question and an answer for each problem.
  
  Label: multiple-question-and-answer