Angle-Arc Theorem

[neno89]

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

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

 

[Guest]

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

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

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 Ainsley :twisted:

 

[pka]

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.

 

[soroban]

Hello, neno89!

Could AB be a diameter?

If so, the proof is very easy.


pka: I believe you proved: $\displaystyle \Delta ADC$ ~ $\displaystyle \Delta ABC$

 

[pka]

pka: I believe you proved: $\displaystyle \Delta ADC$ ~ $\displaystyle \Delta ABC$

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
$\displaystyle A \leftrightarrow D,\quad B \leftrightarrow A,\quad C \leftrightarrow C$.

 

[neno89]

I'm sorry, the conclusion should be triangle ADC is similar to triangle BDA

 

[pka]

I'm sorry, the conclusion should be triangle ADC is similar to triangle BDA

BUT that is FALSE!

 

[neno89]

Hello, neno89!

Could AB be a diameter?

If it was AB is a diameter than what is the next step?

 

[soroban]

Hello, neno89!

If it was AB is a diameter than what is the next step?

                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 $\displaystyle \Delta ADC$