Geometric proof

garfield · **Posted:** Fri, 27 Jan 2012 02:29:52 UTC

Here is one that is puzzling me...

The Question

Prove that angle ACB = 90 degrees (Can someone please tell me how to do degrees signs in latex please?)

Code:

              C
              *
           *  |  *
         *    |    *
       *      |      *
     *        |        *
   *  a       |       b  *
A*____________|____________*B
              O

What I have worked out so far...

1) Triangle ACO

and triangle BCO

are congruent

2) Both of the above triangles are isosceles triangles and $\therefore$ $\angle ACO=a$ degrees and $\angle BCO=b$ degrees also.

Can someone please give me hint on something that can start my proof off please? Its rather frustrating as I can sortoff see how it works in my head but cannot put it into a proof...

Thanks heaps

Shadow · **Posted:** Fri, 27 Jan 2012 06:43:42 UTC

garfield wrote:

Here is one that is puzzling me...

The Question

Prove that angle ACB = 90 degrees (Can someone please tell me how to do degrees signs in latex please?)

Code:

              C
              *
           *  |  *
         *    |    *
       *      |      *
     *        |        *
   *  a       |       b  *
A*____________|____________*B
              O

What I have worked out so far...

1) Triangle ACO

and triangle BCO

are congruent

2) Both of the above triangles are isosceles triangles and $\therefore$ $\angle ACO=a$ degrees and $\angle BCO=b$ degrees also.

Can someone please give me hint on something that can start my proof off please? Its rather frustrating as I can sortoff see how it works in my head but cannot put it into a proof...

Thanks heaps

Hold on, how do you know they are congruent now? Which proof of congruence are you using? All I see is that both have two consecutive sides the same, which is not enough to conclude congruence. You'd need, for example, the included angled to be the same to conclude that. Also, degrees in $\LaTeX$ is given by

Code:

[tex]^\circ[/tex]

garfield · **Posted:** Sun, 29 Jan 2012 09:24:17 UTC

Sorry, I dont actualy really know where the congruency bit came from - must have been too late at night! :oops:

I cannot see anything more to prove it, but there must be somewhere????? :confused:

Can someone please give me a hint???

Thanks alot

outermeasure · **Posted:** Sun, 29 Jan 2012 13:40:26 UTC

garfield wrote:

Sorry, I dont actualy really know where the congruency bit came from - must have been too late at night! :oops:

I cannot see anything more to prove it, but there must be somewhere????? :confused:

Can someone please give me a hint???

Thanks alot

What can you say about triangle ACO?

Soroban · **Posted:** Sun, 29 Jan 2012 14:01:50 UTC

Hello, garfield!

Quote:

$AO\,=\,BO\,=\,CO$

$\text{Prove that }\angle ACB = 90^o$

Code:

                      C
                      o
                  * *  *
              *   *     *
          *     *        *
    A o * * * o * * * * * o B
              O

. .

lie on a circle with center

and radius AO.

$\angle ACB$ is inscribed in a semicircle.

Therefore, $\angle ACB = 90^o$

outermeasure · **Posted:** Sun, 29 Jan 2012 14:24:21 UTC

Soroban wrote:

Hello, garfield!

Quote:

$AO\,=\,BO\,=\,CO$

$\text{Prove that }\angle ACB = 90^o$

Code:

                      C
                      o
                  * *  *
              *   *     *
          *     *        *
    A o * * * o * * * * * o B
              O

. .

lie on a circle with center

and radius AO.

$\angle ACB$ is inscribed in a semicircle.

Therefore, $\angle ACB = 90^o$

The question is probably designed to prove semicircular arc subtends right angle, so quoting that is cheating. :lol:

alstat · **Posted:** Mon, 30 Jan 2012 03:03:16 UTC

If you cannot use Soroban's proof, consider this hint:

Find $m\angle ACO + m\angle BCO$ ,

using the fact that
$m\angle ACO + m\angle CAO + m\angle BCO + m\angle CBO = 180^\circ$ .

S.O.S. Mathematics CyberBoard

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