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## Triangle Area Conjectures

From each side of a right triangle, construct a square. Join the vertices of the squares to their neighboring square to create three new triangles. Measure the areas of these triangles and compare to the original triangle. Conjectures? Suppose the original triangle is not a right triangle. Proof?

Given right triangle ABC, construct squares ADEC, CFGB, and BHIA external to the triangle. One side of each square is also a side of the triangle. Join the vertices of the squares to form the triangles BGH, AID and CEF.

Define all four triangles as polygons and measure their areas. Note that area ABC = area BHG. No surprise, since these two are congruent. However note that area AID = area ABC = area CEF. This is a mild surprise. Deform triangle ABC to discover that these equalities are true for all shapes of ABC.

Question: Why is this?

Hint: Express the areas in terms of two sides and the included angle.

## Translation Using Only Euclidean Compass

Problem: Given segment AB and point C, construct, using only a Euclidean (collapsing) compass, the image, C', of C under the translation that takes A to B. Thus one needs the point C' such that AB and CC' are parallel and equal in length.

On an exam, one geometry student did this: Construct the equilateral triangle ACX (with X toward AB) and construct the equilateral triangle ABY (with Y toward C). Using X and Y construct the equilateral triangle XYH.

Comparing the slope and length of AB with that of CH leads to the conjecture that H = C', the desired point. Dragging C and B seems to confirm the conjecture. How can this be proven?

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