Triangles in Triangles

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

@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]

@AThousandYoung
That is a correct observation.

@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]

angle BAD = CAE
AB = AC, AD = AE
since we have side-angle-side the same:
ABD ~ ACE, so it's always 60*

@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?

@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?

@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)

@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

@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

@forkedknight said
angle BAD = CAE
AB = AC, AD = AE
since we have side-angle-side the same:
ABD ~ ACE, so it's always 60*

@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!

@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