1. R
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    01 Feb '20 01:192 edits
    Begin with an equilateral triangle whose vertices are labeled "A,B,C" ( clockwise from bottom left)

    Place another ( smaller ) equilateral triangle at "A" with vertices "A,D,E" such that "D" lies on line segment "BC", and line segment "DE" intersects line segment "AC" at "P".

    What is angle "ACE"?


    I'm getting hung up on this, not sure what obvious relationship I'm missing. Look forward to seeing the different proofs, and then discussing why my attempt fell short.

    Reveal Hidden Content
    I was hoping to prove that angle DAC was equivalent to angle DEC, proving triangles ADP and EPC were similar? I fell short of finding enough independent equations to achieve this.
  2. Standard memberAThousandYoung
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    01 Feb '20 03:20
    @joe-shmo said
    Begin with an equilateral triangle whose vertices are labeled "A,B,C" ( clockwise from bottom left)

    Place another ( smaller ) equilateral triangle at "A" with vertices "A,D,E" such that "D" lies on line segment "BC", and line segment "DE" intersects line segment "AC" at "P".

    What is angle "ACE"?


    I'm getting hung up on this, not sure what obvious relationship I'm m ...[text shortened]... nd EPC were similar? I fell short of finding enough independent equations to achieve this. [/hidden]
    It seems as though D could lie anywhere along BC
  3. R
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    01 Feb '20 03:381 edit
    @AThousandYoung
    That is a correct observation.
  4. R
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    01 Feb '20 13:462 edits
    @joe-shmo said
    Begin with an equilateral triangle whose vertices are labeled "A,B,C" ( clockwise from bottom left)

    Place another ( smaller ) equilateral triangle at "A" with vertices "A,D,E" such that "D" lies on line segment "BC", and line segment "DE" intersects line segment "AC" at "P".

    What is angle "ACE"?


    I'm getting hung up on this, not sure what obvious relationship I'm m ...[text shortened]... nd EPC were similar? I fell short of finding enough independent equations to achieve this. [/hidden]
    Addendum:

    Connect points "C" and "E" with a line segment to form angle "ACE".
  5. Standard memberforkedknight
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    02 Feb '20 05:48
    angle BAD = CAE
    AB = AC, AD = AE
    since we have side-angle-side the same:
    ABD ~ ACE, so it's always 60*
  6. R
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    02 Feb '20 14:04
    @forkedknight
    That would be correct!

    I was obviously blinded by poor geometry technique in the first place. Well done.

    Follow up: Since your probably better at this, do you see a way to directly show that angle DAP = PEC? Proving triangles DAP and PEC are similar?
  7. Standard memberforkedknight
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    02 Feb '20 16:082 edits
    @joe-shmo said
    @forkedknight
    That would be correct!

    I was obviously blinded by poor geometry technique in the first place. Well done.

    Follow up: Since your probably better at this, do you see a way to directly show that angle DAP = PEC? Proving triangles DAP and PEC are similar?
    Well, we know angle ADP = ACE = 60*
    We know angle APD = EPC because they are opposing at P
    180 - ADP - APD = 180 - ACE - EPC
    therefore, angle DAP = PEC

    Showing two angles the same is enough to know two triangles are similar
  8. R
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    02 Feb '20 17:272 edits
    @forkedknight

    I think you might be misunderstanding me. The objective was to determine angle ACE, so initially we don't know angle ADP = EPC = 60.

    It can be shown that triangle APE is similar to triangle PDC directly by equivalent angles. So, basically I'm asking the question: Is it possible to show ( or is it perhaps necessary ) that the remaining two triangles in the quadrilateral ADCE ( namely ADP and EPC ) are similar. Then somehow work out that angle DAP = PEC, thus proving by equivalence angles ADP = PCE = 60?

    Your proof seems to be the best route, I'm just trying to gage how far away I was from actually finding a solution in the method outlined above (was it right around a corner or infinitely far away)
  9. Standard memberforkedknight
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    02 Feb '20 20:354 edits
    @joe-shmo

    I see what you mean
    By similarity, the ratio of lengths AP : DP = PE : PC
    Therefore DP : PE = AP : PC
    That's enough info to prove ADP ~ ECP
  10. R
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    03 Feb '20 01:072 edits
    @forkedknight said
    @joe-shmo

    I see what you mean
    By similarity, the ratio of lengths AP : DP = PE : PC
    Therefore DP : PE = AP : PC
    That's enough info to prove ADP ~ ECP
    I was trying to show similarity between triangles ADP and EPC using known similarity between triangles APE and DPC. So, DP/AP = CP/EP, and the similarity falls out algebraically through EP/AP=CP/DP and a common angle at "P" (Side-Side-Angle).

    So something seems a little flipped from my notation as "By similarity, the ratio of lengths AP : DP = PE : PC" seems to show similarity between traingles ADP and EPC immediately, making your following statement "Therefore DP : PE = AP : PC" redundant? But yes, that seems to be what I was looking for. Thanks for your solutions!
  11. Standard memberAThousandYoung
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    03 Feb '20 03:001 edit
    @forkedknight said
    angle BAD = CAE
    AB = AC, AD = AE
    since we have side-angle-side the same:
    ABD ~ ACE, so it's always 60*
    How do you know <BAD=<CAE?

    Never mind, I see now

    <BAD+<DAC = <DAC+<CAE = 60
    <BAD=<CAE = 60-<DAC
  12. Standard memberforkedknight
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    03 Feb '20 03:31
    @joe-shmo said
    So something seems a little flipped from my notation as "By similarity, the ratio of lengths AP : DP = PE : PC" seems to show similarity between traingles ADP and EPC immediately, making your following statement "Therefore DP : PE = AP : PC" redundant? But yes, that seems to be what I was looking for. Thanks for your solutions!
    Yes, you are correct, the therefore is redundant.

    I was trying to say that the ratios of the sides of triangles APE and DPC are same by similarity, and the ratios of the sides of triangles DAP and CPE are the same since they share two of the same sides, so therefore must also be similar.
  13. Standard memberAThousandYoung
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    03 Feb '20 03:33
    @forkedknight said
    Well, we know angle ADP = ACE = 60*
    We know angle APD = EPC because they are opposing at P
    180 - ADP - APD = 180 - ACE - EPC
    therefore, angle DAP = PEC

    Showing two angles the same is enough to know two triangles are similar
    Nice
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