**ABC is an acute-angled triangle and P is the foot of the perpendicular dropped from A to BC. If AB + BP = PC, then ∠ABC = 2∠BCA.**

(The converse is also true – it can be proven in an entirely similar manner)

**Proof**

A bit of trig with double angle formulae can do this fairly quickly but I’ll give a synthetic proof that is also quick (and of course more elegant). Let the reflection of B about P be B’.

Then BP = B’P, and AP is the perpendicular bisector of BB’ so triangle ABB’ is isosceles with AB = AB’.

AB + BP = PC = B’P + B’C = BP + B’C

AB = B’C so AB’ = B’C, which makes triangle AB’C isosceles.

∠ABC = ∠AB’B as triangle ABB’ is isosceles.

∠AB’B = ∠BCA + ∠B’AC (by exterior angle of triangle)

But ∠BCA = ∠B’AC as AB’C is isosceles so ∠AB’B = 2∠BCA

So ∠ABC = 2∠BCA

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