perplexus.info :: Geometry : Splitting Hypotenuse

Splitting Hypotenuse (Posted on 2011-04-03)

Two rays from the right-angled vertex of an isosceles right triangle split the hypotenuse into three sgments.

Prove that the three segments can be used to form a right triangle (the middle segment as the hypotenuse) if and only if the angle between the rays is 45°.

Submitted by Bractals

Rating: 4.0000 (2 votes)

Solution: (Hide)

Let ABC be the isosceles right triangle with right angle at vertex C. Let the rays from C intersect the hypoteneuse at points D and E such that we have the collinear points A, D, E, and B in that order.

Construct point F on the opposite side of BC from point E such that triangles CBF and CAD are congruent. Construct line segment EF.
/EBF = /EBC + /CBF = /EBC + /CAD = 45° + 45° = 90° |EF|² = |EB|² + |BF|² = |EB|² + |AD|²
Consider triangles DCE and FCE: sides DC and FC are congruent and side CE is congruent to itself.
|DE|² = |AD|² + |EB|² <==> |DE| = |FE| <==> ΔDCE and ΔFCE are congruent <==> /DCE = /FCE <==> /DCE = /FCB + /BCE <==> /DCE = /ACD + /ECB <==> /DCE = 90° - /DCE <==> /DCE = 45°

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Another approach (spoiler)	Harry	2011-04-05 19:13:08
Possible solution	broll	2011-04-04 07:06:30

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