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#1 h Sep 7, 2004, 11:00 PM • 2 Y

Y by Adventure10, Mango247

Let ABCDE be a convex pentagon such that each diagonal is parallel to at least one side of the pentagon. Also it is Known that BC=CD=DE. Prove it is regular!

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#2 h Sep 8, 2004, 7:25 PM • 2 Y

Y by Adventure10, Mango247

Two things are clear: $BCDE$ is an isosceles trapezoid, and $A$ lies on the perpendicular bisector of $BE$ (or $CD$ ; it's the same perpendicular bisector). We have $AD\|BC,\ AE\|BD$ , so $\angle CBD=\angle DAE$ . From the additional fact that $CD=DE$ , we find $A$ to be on the circumcircle of $BCDE$ (which is cyclic because it's an isosceles trapezoid). This means that, for example, $ACDE$ is also an isosceles trapezoid, so $AE=CD$ , and this pretty much ends the proof.

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#3 h Sep 9, 2004, 9:01 AM • 2 Y

Y by Adventure10, Mango247

Grobber, your proof is very nice. Actually, the problem is funny: it looks like something one has to think about, but it turns out to be a simple angle chase... So here is my proof, really no brain needed:

Since "each diagonal is parallel to at least one side of the pentagon", the diagonal AC must be parallel to the side DE (in fact, it is parallel to at least one side of the pentagon, but it can't be parallel to AB, to BC, to CD, or to AE, hence it must be parallel to DE). Similarly, the diagonals BD, CE, DA, EB are parallel to the sides EA, AB, BC, CD, respectively.

The quadrilateral BCDE is an isosceles trapezoid (in fact, it is a trapezoid since BE || CD, and since BC = CD = DE, it has equal legs, so that it is either an isosceles trapezoid or a parallelogram, but it cannot be a parallelogram, since else we would have BC || DE, what, together with AC || DE, would yield BC || AC, impossible).

Now call t = < DBE. Then,

t = < CDB (since t = < DBE, and < DBE = < CDB since EB || CD);
t = < DCE (since t = < CDB, and < CDB = < DCE since the quadrilateral BCDE is an isosceles trapezoid);
t = < CEB (since t = < DBE, and < DBE = < CEB since the quadrilateral BCDE is an isosceles trapezoid);
t = < BEA (since t = < DBE, and < DBE = < BEA since BD || EA);
t = < ABE (since t = < CEB, and < CEB = < ABE since CE || AB);
t = < CBD (since t = < CDB, and < CDB = < CBD since the triangle BCD is isosceles);
t = < ADB (since t = < CBD, and < CBD = < ADB since DA || BC);
t = < DAE (since t = < ADB, and < ADB = < DAE since BD || EA).

Now, we have t = < DBE = < DCE = < DAE. Hence, the points A, B, C, D, E lie on one circle. Each of the chords AB, BC, CD, DE, EA has the same chordal angle in this circle, namely the angle t (this is because t = < ADB = < CDB = < CBD = < DCE = < ABE). Hence, the pentagon ABCDE is regular. Qed..

Darij

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