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Side of a triangle (geometry)

  1. Aug 26, 2015 #1
    1. The problem statement, all variables and given/known data
    As shown in the diagram below, the shape consists of a square and a circle with centre Q. Given that QM = 3 cm, prove that MN = 6 cm.

    Known data:
    -- triangles APB and BQC are congruent
    -- angle BMC = 90
    -- triangles BMQ and BNA are similar and right-angled

    Doubt.JPG

    2. Relevant equations
    3. The attempt at a solution


    This is a part of a structured question and I've already proved all the data listed above. I also know that AN = 6 cm as the triangles are similar, and I think that the solution might have something to do with triangle AMN being isosceles (but I can't prove that). There's probably something obvious I'm missing, so a nudge in the right direction would be greatly appreciated.
     
  2. jcsd
  3. Aug 26, 2015 #2

    RUber

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    Can you show that there is a point X such that ANMX is a square?
     
  4. Aug 26, 2015 #3
    Thank you! I think something like this would suffice: Let AX be the line parallel to MN, such that X lies on the extension of CQ. As we know ∠NMQ = 90°, ∠AXM =90° as they are interior angles between parallel lines. Since ∠MNA and ∠XAN are 90° as well, ANMX is a square, and AN = NM. Therefore MN is 6 cm.

    Thanks once again.
     
  5. Aug 26, 2015 #4

    RUber

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    I think that your argument only proves that ANMX is a rectangle. I may have steered you wrong...I do not see a good way to make the conclusion along those lines.
    What about this:
    triangles APB and BQC are congruent, so AP = AQ.
    So you should be able to show that QMB = NPA.
    Then use the isosceles rule to say BN = NM.
    And I think you should be able to conclude the result you are looking for.
     
  6. Aug 26, 2015 #5

    SammyS

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    You should be able to do this using similar triangles.

    From the result that AN = 6cm, you can show that ΔAMN is isosceles.
     
  7. Aug 26, 2015 #6
    What about this:

    As ΔAPB is congruent with ΔBQC, AP = QB
    ΔANB is congruent with ΔBMC by AAS rule since ∠ANB = ∠BMC = 90°, ∠MCB = ∠NBA, AB = BC
    Hence, AN = BM, the third side of the congruent triangles shown. As AN = 6 cm, BM = 6 cm
    As ΔBMQ and ΔBNA are similar (scale factor 2), BN = BM*2 = 12 cm, and MN = BN - BM = 6 cm.
     
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