Now let's prove the converse theorem: If the median to a side is equal to half that side, then the triangle is a right triangle. Nice and simple - thanks, Ben! Converse Theorem And |CD| it is equal to half of |AB| it since it is one of the equal legs in both the isosceles triangles we created! Using the transitive property of equality, if |DC|=|DB| and |DC|=|DA|, then |DB|=|DA|.īut if |DB|=|DA|, then by definition CD is the median to the hypotenuse. So m∠ACD=m∠BAC=90°-α, and ΔDCB is also an isosceles triangle and |DC|=|DA|. And since m∠ACB=90° and we constructed ∠DCB≅∠DBC=α, then m∠ACD=90°-α. If ∠DBC=α, and m∠ACB=90°, then by the sum of angles in a triangle theorem, m∠BAC=90°-α. Since we constructed ∠DCB≅∠DBC, by the converse base angles theorem, ΔDCB is isosceles and |DC|=|DB|. 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 length of the median to the hypotenuse is half the length of the hypotenuse, as we'll show in this proof. The median of a triangle is a line drawn from one of the vertices to the mid-point of the opposite side.
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