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Math Help - Help with these two geometry!

  1. #1
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    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.
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  2. #2
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    Hello xxravenxx
    Quote Originally Posted by xxravenxx View Post
    ... 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)

    Grandad
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  3. #3
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    Sorry but can you draw me a diagram of what this looks like? I'm a bit clueless.
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  4. #4
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    Hello xxravenxx
    Quote Originally Posted by xxravenxx View Post
    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:
    Quote Originally Posted by xxravenxx View Post
    ...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)

    Grandad
    Attached Thumbnails Attached Thumbnails Help with these two geometry!-untitled.jpg  
    Last edited by Grandad; October 21st 2009 at 01:36 AM. Reason: Add solution to qu 13
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