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Math Help - Triangle. Equality of angles

  1. #1
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    Triangle. Equality of angles

    Point M is the middle of the side BC of an acute-angles triangle ABC. Point K is on the side BC and . On section AK there is a point E and . Prove that .

    (Symbol "<" means an angle)

    I would be very grateful for any help
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  2. #2
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    Quote Originally Posted by PaulinaAnna View Post
    Point M is the middle of the side BC of an acute-angles triangle ABC. Point K is on the side BC and . On section AK there is a point E and . Prove that .

    (Symbol "<" means an angle)
    This is another tricky problem. The thing I found hard was to know how to make use of the fact that M is the midpoint of BC. Eventually, it occurred to me that this means that the triangles BAM, CAM have the same area (equal bases, same vertical height).

    Denote by \angle A,\ \angle B,\ \angle C the three angles of the triangle. Let \alpha = \angle BAM = \angle KAC and let \beta = \angle BAK = \angle MAC.

    From B, draw a line BX to meet AK at X, such that \angle EBX = \angle B. Then the triangles ABC, EBX are similar (because \angle BEX = \angle A and \angle EBX = \angle B), and so \dfrac{BE}{EX} = \dfrac{AB}{AC}.

    Also, \angle BXA = \angle C and hence the points A, B, C, X are concyclic (having equal angles in the same segment). Therefore \angle BXC = \angle B + \angle C (opposite angles of a cyclic quadrilateral are supplementary), and so \angle EXC = \angle B. Also, \angle XBC = \angle KAC = \alpha and \angle XCB = \angle BAK = \beta (angles in the same segment).

    The area of triangle BAM is \frac12AB.AM\sin\alpha and the area of triangle CAM is \frac12AC.AM\sin\beta. Since these are equal, it follows that \dfrac{AB}{AC} = \dfrac{\sin\beta}{\sin\alpha}.

    Now look at the (vertical) distance from X to BC. This is equal to BX\sin\alpha (since \angle XBC = \alpha). It is also equal to CX\sin\beta (since \angle XCB = \beta). Therefore \dfrac{BX}{CX} = \dfrac{\sin\beta}{\sin\alpha} = \dfrac{AB}{AC} = \dfrac{BE}{EX}.

    Putting together all the above facts, it follows by the RAR rule for similarity that the triangles EBX, EXC are similar. In particular, \angle KEC = \angle BEX = \angle A, as required.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Also, \angle BXA = \angle C
    Where do we know this from?
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  4. #4
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    Quote Originally Posted by Glyper View Post
    Quote Originally Posted by Opalg View Post
    Also, \angle BXA = \angle C
    Where do we know this from?
    This follows from the fact that the triangles ABC and EBX are similar. The angle at X in triangle EBX corresponds to the angle at C in triangle ABC.
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  5. #5
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    Oh, thank you very much! I am stuck again, though:

    Quote Originally Posted by Opalg View Post
    the points A, B, C, X are concyclic (having equal angles in the same segment).
    How do we know they're all concyclic and what equal angles in "the same segment" did you mean?
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