![]() In right triangle ΔABC, let's construct a line CD, that splits the right angle ∠ACB into 2 angles, ∠ACD and ∠DCB, such that ∠DCB≅∠DBC, and call that angle α. ![]() Instead, it uses angle construction and some simple angle math to show that the median creates two isosceles triangles. And indeed, one of my regular readers sent me the following solution, which is quite elegant, and does not rely at all on the midsegment theorem. ![]() One of the fun things about these proof problems is that often there is more than one way to approach and prove the theorem. (14) CD= ❚B //(12), (13), substitution Another way to prove this (13) DB= ❚B // Given, CD is the median to the hypotenuse (12) CD=DB // Corresponding sides in congruent triangles (CPCTC) (10) DE=DE //common side, reflexive property of equality (9) ∠DEB ≅ ∠DEC //(7),(8), definition of congruent angles (6) m∠ACE=90° //given, ΔABC is a right triangle (5) ∠DEB ≅ ∠ACE // corresponding angles in two parallel lines intersected by a transversal line (CE) (4) DE||AC // triangle midsegment theorem (3) DE is a midsegment //(1), (2), Definition of a midsegment (1) AD=DB //given, CD is the median to the hypotenuse ![]() So m∠DEC=90°, too, as it forms a linear pair with ∠DEB.Īnd we can now prove the triangles ΔDCE and ΔDBE are congruent using the Side-Angle-Side postulate, with the result that CD=DB as the corresponding sides, just as we need to show.Īs a result, we also see that the median to the hypotenuse creates two isosceles triangles, ΔDCB and ΔDAC, where DA=DC=DB. From this, we know ∠DEB ≅ ∠ACE (as corresponding angles) and they are both right angles. Now, D is the midpoint of the hypotenuse, and E is the midpoint of the leg CB, so DE is a midsegment, and using the triangle midsegment theorem, we know DE||AC. This will have the advantage of creating two triangles where the segments we want to prove are equal are corresponding sides and having two sides we know are equal (CE and EB, as E is the midpoint). Let's construct such triangles, by connecting point D (the midpoint of the hypotenuse) with the middle point of CB. So we will need to construct new triangles in which the segments we want to prove are equal are corresponding sides. In which the two triangles can't possibly be congruent. And while it is never a good idea to rely on the drawing to make conclusions, we can imagine a right triangle that looks like this : It might be tempting to try to use the existing triangles created by the median (ΔACD, ΔDCB), but a quick look at the drawing shows us that can't be right. Show that AD=DC BD=DC StrategyĪs we need to show that a couple of line segments are equal (AD=DC BD=DC) the tool we'll use is triangle congruency. In the right triangle ΔABC, line segment CD is the median to the hypotenuse AB. In the case of a right triangle, the median to the hypotenuse has the property that its length is equal to half the length of the hypotenuse. The median of a triangle is a line drawn from one of the vertices to the mid-point of the opposite side. ^ Leung, Kam-tim and Suen, Suk-nam "Vectors, matrices and geometry", Hong Kong University Press, 1994, pp.In today's geometry lesson, we will prove that in a right triangle, the median to the hypotenuse is equal to half the hypotenuse.^ Benyi, Arpad, "A Heron-type formula for the triangle", Mathematical Gazette 87, July 2003, 324–326.^ Boskoff, Homentcovschi, and Suceava (2009), Mathematical Gazette, Note 93.15.^ Posamentier, Alfred S., and Salkind, Charles T., Challenging Problems in Geometry, Dover, 1996: pp.^ Sallows, Lee, " A Triangle Theorem" Mathematics Magazine, Vol.E., "Halving a triangle," Mathematical Gazette 56, May 1972, 105-108. "Medians and Area Bisectors of a Triangle". CRC Concise Encyclopedia of Mathematics, Second Edition. The lengths of the medians can be obtained from Apollonius' theorem as: If the two triangles in each such pair are rotated about their common midpoint until they meet so as to share a common side, then the three new triangles formed by the union of each pair are congruent. In 2014 Lee Sallows discovered the following theorem: The medians of any triangle dissect it into six equal area smaller triangles as in the figure above where three adjacent pairs of triangles meet at the midpoints D, E and F. (Any other lines which divide the area of the triangle into two equal parts do not pass through the centroid.) The three medians divide the triangle into six smaller triangles of equal area.Ĭonsider a triangle ABC. Each median divides the area of the triangle in half hence the name, and hence a triangular object of uniform density would balance on any median.
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