AoPS - Problem

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Problem

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Tags: geometry, Squares, sqauare, equal segments, perpendicular

parmenides51

04.06.2020 15:49

Let $ABCD$ be a square and $E$ a point on the side $(CD)$ . Squares $ENMA$ and $EBQP$ are constructed outside the triangle $ABE$ . Prove that: a) $ND = PC$ b) $ND\perp PC$ .

gghx

05.06.2020 05:41

This is a nice question, does anyone have a solution?

Gogobao

05.06.2020 06:10

Let

$A = (0, 1), B = (0, 0), C = (1, 0), D = (1, 1), E = (1, y)$ Then

$N = (2-y, y+1), P = (1+y, y-1)$

$ND^2 = (1-y)^2 + y^2 = 2y^2+2y+1, PC^2 = y^2 + (y-1)^2 = 2y^2+2y+1$ , so

$ND = PC$ Vector

$ND = [y-1, -y]$ and Vector

$PC = [-y, 1-y]$

$[y-1, -y] \cdot [-y, 1-y] = 0$ , so

$ND \perp PC$

There's probably some synthetic solution but coord bashing worked.