Equal Segments (Posted on 2013-03-20)

Let M be the circumcenter of ΔABC ( ∠B = 90° ),
    E the point on altitude BD of ΔABC such that 
      ∠AEC = 135°,
    F the point such that BDCF is a rectangle. and
    G the intersection of line segment MF and the
      circumcircle of ΔABC.

Prove that |DE| = |FG|.

	Submitted by Bractals
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Let a = |BC|, b = |CA|, c = |AB|, x = |DE|, and y = |FG|. From the similar right triangles ABC, ADB, and BDC we get |DA| = c2/b, |DB| = ac/b, |DC| = a2/b. |FG| + |GM| = |FM| ==> |FG|2 + 2|GM||FG| + |GM|2 = |FC|2 + |CM|2 ==> |FG|2 + 2|GM||FG| = |DB|2 ==> y2 + by = (ac/b)2 ==> b2y2 + b3y = a2c2. (1) -1 = tan(135°) = tan(∠AEC) = tan(∠AED + ∠DEC) tan(∠AED) + tan(∠DEC) = ------------------------ 1 - tan(∠AED)tan(∠DEC) (c2/bx) + (a2/bx) = ------------------- 1 - (c2/bx)(a2/bx) b3x = ------------ ==> b2x2 - a2c2 b2x2 + b3x = a2c2. (2) Combining (1) and (2) gives b2x2 + b3x = b2y2 + b3y ==> b2(x + y + b)(x - y) = 0 ==> x = y ==> |DE| = |FG|. QED

	Subject	Author	Date
	Different approach	Harry	2013-03-21 15:36:41
	Solution	Jer	2013-03-20 22:18:31