Angle-Arc Theorem

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neno89

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Oct 25, 2005
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Hey! Can someone help me solve this problem? I am confused and not sure what to do. Thank you for your help!

212cx.png


Given: Segment AC is tangent to circle O at A.
Conclusion: Triangle ADC is similar to triangle BDA
 
neno89 said:
Hey! Can someone help me solve this problem? I am confused and not sure what to do. Thank you for your help!

212cx.png


Given: Segment AC is tangent to circle O at A.
Conclusion: Triangle ADC is similar to triangle BDA
Click to expand...

Hey neno89!! have you tried just writing out the proof?
example :
1) Segment AC is tangent to circle O @ A 1) Given
2) Etc
3)etc...........................................................................................................

Im not the best at geometry but i know that writing the entire problem out always helps me :wink: Also, have you tried the bisector of an angle? (what about proving the triangles?)
It's not much but i hope this helps 8-) Ainsley :twisted:
 
Given: Segment AC is tangent to circle O at A.
Conclusion: Triangle ABC is similar to triangle DAC.
I think that you made a mistake in writing the conclusion.
PROOF:
Because both intercept arc AD we have \(\displaystyle \angle ABC\tilde = \angle DAC\) and \(\displaystyle \angle C\tilde = \angle C\).
Two angles being congruent means the triangles are similar.
 
Hello, neno89!

Could AB be a diameter?

If so, the proof is very easy.


pka: I believe you proved: \(\displaystyle \Delta ADC\) ~ \(\displaystyle \Delta ABC\)
 
soroban said:
pka: I believe you proved: \(\displaystyle \Delta ADC\) ~ \(\displaystyle \Delta ABC\)
Click to expand...
No, actually I did correct the conclusion neno89 wrote.
Moreover, what I did prove is \(\displaystyle \Delta ABC \approx \Delta DAC\)! Also note the angle correspondence is
\(\displaystyle A \leftrightarrow D,\quad B \leftrightarrow A,\quad C \leftrightarrow C\).
 
I'm sorry, the conclusion should be triangle ADC is similar to triangle BDA
 
Hello, neno89!

If it was AB is a diameter than what is the next step?
Click to expand...
Code: 
                A             C
              * * * - - - - *
          *     |  \  *    /
        *       |     \ * /
       *        |        * D
                |       /
      *         |      /  *
      *        O*     /   *
      *         |    /    *
                |   /
       *        |  /     *
        *       | /     *
          *     |/    *
              * * *
                B
Since \(\displaystyle AOB\) is a diameter, \(\displaystyle \angle ADB\,=\,\angle ADC\,=\,90^o.\)

In right triangle \(\displaystyle ADB,\;\angle ABD\) is measured by \(\displaystyle \,\frac{1}{2}\text{ arc}AD.\)

In right triangle \(\displaystyle ADC,\;\angle CAD\) is measured by \(\displaystyle \,\frac{1}{2}\text{ arc}AD.\)

Hence: \(\displaystyle \,\angle ABD\,=\,\angle CAD.\)

Therefore: \(\displaystyle \,\Delta ADB\) is similar to \(\displaystyle \Delta ADC\)
 
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