A chord $AB$ of one of two concentric circles at intersect each other at $C$ and $D$. We have to prove, $AC=BD$.

I am not sure what this question means by 'intersect each other', but if I am correct, we can assume that $AB$ is the chord of the outer circle that intersects the inner one at $C$ and $D$. I proved in one of my exams that if their centre is at $O$, triangles $\triangle ACO$ and $\triangle BDO$ are congruent by SAS. Thus $AC=BD$, proving the theorem. The teacher, however, gave me zero marks and left only one comment, 'Wrong derivation'. I do not even know what he means.So what is wrong with my proof?

ADDED: For a more detailed explanation of what I did, I joined $O$ with $A$, $B$, $C$ and $D$. I argued that since $OC$ and $OD$ are equal, angles $\angle OCD$ and $\angle ODA$ are equal, and thus angles $\angle OCA$ and $\angle ODB$ are equal. By similar reasoning, $\angle OAC$ and $\angle OBD$ must be equal. Therefore, the remaining angles must be equal. Therefore, by SAS, triangles $\triangle ACO$ and $\triangle DBO$ are congruent. Thus $AC$ and $BD$ are equal.

  • $\begingroup$ Perhaps the text is intended to be "A chord $AB$ of one of two concentric circles intersects the other at $C$ and $D$." Briefly: "A chord of one circle intersects the other circle." So, your reading seems to be correct. (I haven't checked your proof, though.) $\endgroup$ – Blue Aug 26 '13 at 3:47
  • $\begingroup$ @Blue,so I was correct all along? $\endgroup$ – rah4927 Aug 26 '13 at 5:30
  • $\begingroup$ I just ran through your proof again (I mean ... for the first time ;). It looks good to me ... except for a minor-minor quibble: you mean "$\triangle ACO$ and $\triangle BDO$ by SAS" in your first paragraph. (The triangles should be named in the same order.) $\endgroup$ – Blue Aug 26 '13 at 5:36
  • $\begingroup$ The only thing I really notice that that you worked slightly too hard: once you have the two angle congruences, you can say that the triangles are congruent by SAA (or AAS); going the extra step for SAS is unnecessary. (Then again, the whole reason SAA works is because that extra step makes it equivalent to SAS.) $\endgroup$ – Blue Aug 26 '13 at 5:38
  • $\begingroup$ @Blue,yes,I find remembering the SAA criterion hard for some reason.Ty by the way. $\endgroup$ – rah4927 Aug 26 '13 at 5:42

Credits go to Blue for checking the proof.I am merely trying to remove the question from the unanswered queue.My way of showing congruency through SAS does work.However,as pointed out by Blue,I simply could have taken the SAA route,which is one step short.Should anyone else have any advice on my proof,I shall be quite happy to listen.

EDIT: Okay,here is what the teacher meant by 'the correct derivation':

We drop a perpendicular OP on the chord.So the perpendicular bisects the chord(s) at P.Therefore,

AP=PB. . . . . . (1)

CP=DP. . . . . .(2)

Subtracting (2) from (1) gives us the required result.

  • 1
    $\begingroup$ @Blue,okay then. $\endgroup$ – rah4927 Aug 27 '13 at 17:42

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