Question

4. In a tight angle triangle length of median of hypolenuse 19cm.
Find the length of hypotenuse​

Answer 1

Step-by-step explanation:

(1) AD=DB //given, CD is the median to the hypotenuse

(2) CE=EB //construction

(3) DE is a midsegment //(1), (2) , Definition of a midsegment

(4) DE||AC //triangle midsegment theorem

(5) ∠DEB ≅ ∠ACE // corresponding angles in two parallel lines intersected by a transversal line (CE)

(6) m∠ACE=90° //given, ΔABC is a right triangle

(7) m∠DEB=90° //(5), (6), definition of congruent angles

(8) m∠DEC=180° – 90° = 90° // linear pair

(9) ∠DEB ≅ ∠DEC //(7),(8), definition of congruent angles

(10) DE=DE //common side, reflexive property of equality

(11) ΔDCE ≅ ΔDBE //(2), (9), (10) Side-Angle-Side postulate

(12) CD=DB // Corresponding sides in congruent triangles (CPCTC)

(13) DB= ½AB // Given, CD is the median to the hypotenuse

(14) CD= ½AB //(12) , (13), substitution

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Another way to prove this

One of the fun things about these proof problems is that often there is more than one way to approach and prove the theorem. 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. Instead, it uses angle construction and some simple angle math to show that the median creates two isosceles triangles.

median to hypotenuse

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. Since we constructed ∠DCB≅∠DBC, by the converse base angles theorem, ΔDCB is isosceles and |DC|=|DB|.

Now let’s do some angle math. If ∠DBC=α, and m∠ACB=90°, then by the sum of angles in a triangle theorem, m∠BAC=90°-α. And since m∠ACB=90° and we constructed ∠DCB≅∠DBC=α, then m∠ACD=90°-α. So m∠ACD=m∠BAC=90°-α, and ΔDCB is also an isosceles triangle and |DC|=|DA|. Using the transitive property of equality, if |DC|=|DB| and |DC|=|DA|, then |DB|=|DA|.

But if |DB|=|DA|, then by definition CD is the median to the hypotenuse. 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!

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