Geometry and Trig

skinski43

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Jul 5, 2009
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In a right triangle, the median and altitude drawn to the hypotenuse make an angle congruent to the difference of the acute angles of the triangle.
Can anyone help me with this proof?
 
The triangle formed by the median is an isosceles triangle. Hence two of its angle are equal. The triangles formed by the altitude are both right triangles. They are similar to each other and to the original triangle. Henc e corresponding angles of those triangles are equal. You know the sum of the three angles formed at the right angle by the median and the altitude is 90 degrees. The sum of the acute angles of any right triangle is 90 degrees.
I hope those hints provide enough information for you to get it. If not, show us your work and let us know your thinking. It's relatively straight forward. Don't try to make it too complex.
 
Thank you for your help Loren. I will change my strategy and show you my work.

Skinski43
 
Loren, I'm unable to prove that the triangle formed by the median is isosceles. If I can prove both of those angles are equal the rest of the proof goes well.

skinski43
 
Hello, skinski43!

Prove: in a right triangle, the median and altitude drawn to the hypotenuse
make an angle congruent to the difference of the acute angles of the triangle.
Code:
            C
            *
           *|  *
          *b| a   *
         *  |        *
        *   |           *
       * a  |            b *
    A * - - * - - - - - - - - * B
            D

\(\displaystyle \text{We have: right triangle }ABC\!:\;C = 90^o\:\text{ and altitude }CD.\)

\(\displaystyle \text{Let: }\:a = \angle CAB,\;\;b = \angle CBA\)

\(\displaystyle \text{Altitude }CD\text{ forms three similar right triangles.}\)

\(\displaystyle \text{Hence: }\:\angle ACD = b,\;\;\boxed{\angle BCD = a}\)



Code:
            C
            *
           * * *
          *   * b *
         *     *     *
        *       *       *
       * a       *       b *
    A * - - - - - * - - - - - * B
                  M

\(\displaystyle \text{We have right triangle }ABC\!:\;\;C = 90^o\:\text{ and median }CM.\)

\(\displaystyle \text{Since a right triangle can be inscribed in a semicircle,}\)
. . \(\displaystyle \text{median }CM \:=\:AM \:=\:BM.\)

\(\displaystyle \text{Hence, }\Delta MCB\text{ is isosceles: }\:\angle MBC = \boxed{\angle MCB = b}\)



Code:
            C
            *
           *|* *
          * | *   *
         *  |  *     *
        *   |   *       *
       * a  |    *       b *
    A * - - * - - * - - - - - * B
            D     M

\(\displaystyle \text{Therefore: }\:\angle DCM \:=\:\angle BCD - \angle MCB \;=\;a - b\)

 
skinski43 said:
Loren, I'm unable to prove that the triangle formed by the median is isosceles. If I can prove both of those angles are equal the rest of the proof goes well.

skinski43

If ABC is a right angled triangle - with BC as hypotenuse and O as the midpoint of BC

Then you can draw a circle, through A, B & C, with O as center and BC as diameter. Now you can easily see that AO (the median) is also radius - hence:

AO = BO = OC

Okay now Soroban beat me to it - but I'll leave my redundant answer anyway.....
 
Many thanks to you Loren, and to Subhotosh Kahn for helping me with this problem. I've been working for the past two weeks using 2 different textbooks without success.
I'm sure I will sleep better tonight. Do either of you recommend a book that helps students to learn geometry, especially proofs. I have been using "Geometry for Dummies" by Mark Ryan. This book does emphasize proofs. The other book I used was "Geometry Book I. Planimetry" by A. P. Kiselev. I'll go back to these books to see if either had your explanation and I missed it. It may be my fault, but if you recommend a good learning tool I will try to purchase it. thank you both again, God bless.

Skinski43
 
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