Angle Bisector in Parallelogram
What Is It About?
A Mathematical Droodle

Explanation

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Copyright © 1996-2012 Alexander Bogomolny

The applet illustrates Problem 2 from IMO 2007:

The solution is an adaptation of that by Anton Batominovski (a simpler variant appears elsewhere.)

The condition EF = EG = EC is equivalent to E being the circumcenter of ΔCFG. Thus we shall prove a stronger result:

Assume first that l is the bisector of ∠BAD. This makes ΔABG isosceles so that BG = AB. Since ABCD is a parallelogram we also have CD = AB implying BG = CD.

ΔFCG is also isosceles (and similar to ΔABG.) In particular, EC being its apex altitude and angle bisector,

Bt, since E is the circumcenter of ΔCFG, ΔCEG is isosceles, ∠CGE = ∠ECG making ∠ECF = ∠CGE. The bottom line is in triangles CDE and GBE, EC = EG, CD = BG, and ∠CGE = ∠ECG, from which ΔCDE = ΔGBE. The two triangles are obtained from each other by a rotation around E. This implies the angles between corresponding sides are equal, namely, ∠BCD = ∠BED so that E is inscribed into the circumcircle of BCD.

Now let's prove the converse. Assume E lies on the circumcircle of ΔBCD. By the Simson-Wallace Theorem theorem, the feet of the perpendiculars EK, EL, EM from E to the sides BC, CD, BD are collinear on the Simson line, where K is the midpoint of CG, while L is the midpoint of CF. The homothety with center C and coefficient 2 maps K onto G, L onto F, the line KL onto l, and the diagonal BD onto a parallel line through A. M being on the two lines is mapped to the intersection of their images, which is A. This makes M the midpoint of the diagonal AC and, as a consequence, also the midpoint of the diagonal BD. ΔBED is then isosceles, with BE = DE and ∠DBE = ∠BDE and CM ⊥ BD.

On the other hand, ∠DBE = ∠DCE (as inscribed angles subtended by the same chord). Also, since BCDE is cyclic, ∠BDE is supplementary to ∠BCE as is ∠GCE so that ∠BDE = ∠GCE leading finally to

making CE an angle bisector in ΔCFG. But since E is the circumcenter of ΔCFG it also becomes the altitude and the median from C which means that ΔCFG is isosceles as are triangles ADF and ABG. All three are similar and we finally get ∠DAG = ∠BAG.

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Copyright © 1996-2012 Alexander Bogomolny