This property is quite well known- If P is a point on the minor arc BC of the circumcircle of an equilateral triangle ABC, then PB+PC=PA.

I discovered the following- If P is a point in the plane of equilateral triangle ABC such that PB+PC=PA, then P is a point on the minor arc BC.

Try to prove this using TONCAS. Hint- A lawyer has a point to make.

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TopNewestwell,I think i haVE GOT ITlet P not be on minor arc BC..take point J in the interior of tr.ABC such that BJ=BP and ang.PBJ=60. so,tr. ABJ is congruent to tr.CBP by SAS congruence criterion..so AJ=PC..JP=BJ=BP(tr.BJP is equilateral)Now consider tr.AJP by triangle inequality we get AP is less than AJ+JP.but AJ+JP=PB+PC...AP=PB+PC...hence contradiction...hence proved!! is this correct? – Neeraja Kirtane · 1 year, 9 months ago

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– Pranav Kirsur · 1 year, 9 months ago

Yes,this is correct. I like your construction. I used Fermat Point to get a contradiction.Log in to reply

– Neeraja Kirtane · 1 year, 9 months ago

What did you do exactly using fermat's point?Log in to reply

– Pranav Kirsur · 1 year, 9 months ago

I have considered a case where P is not on the minor arc BC. I have used the property that the Fermat point minimises the sum of the distances from the vertices of a triangle.Log in to reply

– Neeraja Kirtane · 1 year, 9 months ago

hmm..but how does point P come in this context?Log in to reply

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– Pranav Kirsur · 1 year, 10 months ago

I have already shown it to sir.Log in to reply

1000 followers problem. – Nihar Mahajan · 1 year, 9 months ago

Try myLog in to reply

– Nihar Mahajan · 1 year, 10 months ago

Ok , then no problem!Log in to reply