![]() 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. ![]() In today's geometry lesson, we will prove that in a right triangle, the median to the hypotenuse is equal to half the hypotenuse.
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