In the following diagram, PT and PU are tangents. Prove that MUPT is a cyclic quadrilateral.


In order to use the $\text{(ext $\angle = $ int opp $\angle$)}$ rule:

$\widehat{U_4} = \widehat{T_2}\quad\text{(tan chord)}$

but now I can't prove that $\widehat{T_1} = \widehat{U_3}$.

I have drawn this in GeoGebra, so I can graphically see that it is true.


A more complete version of qsmy's answer is this*:

Let the intersection of the rays extending from $PU$ and $TW$ be $Q$.

$\angle UPM=\angle QUW\quad\text{(corr. $\angle$'s, $PM||UW$)}$

$\angle QUW=\angle UTW\quad\text{(tan chord)}$

$\therefore \angle UPM = \angle UTW$

$\therefore MUPT \text{ is a cyclic quad}\quad\text{(line segment subtends $= \angle$'s)}$

To make the last line clear, the lemma used is this: "If a line segment subtends equal angles at two other points on the same side of the line segment, then these four points are concyclic."

*qsmy and others did not want that answer to be edited.


Let the ray where $PU$ is be $PQ$. It is actually simply $∠UPM=∠QUW=∠UTW$.

  • $\begingroup$ I understand what you have written, but can you please complete the proof? (To prove that $MUPT$ is a cyclic quadrilateral) $\endgroup$ – ahorn Apr 25 at 8:08
  • $\begingroup$ That equality can immediately deduce that the four points are concyclic $\endgroup$ – qsmy Apr 25 at 8:31
  • $\begingroup$ Please refer to which lemma you are talking about. $\endgroup$ – ahorn Apr 25 at 8:42
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    $\begingroup$ you can refer to this: youtube.com/watch?v=zP5Fta_Dch0 $\endgroup$ – qsmy Apr 25 at 9:13
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    $\begingroup$ @ahorn My suggestion is to write up your own answer rather than editing this one. You can thank the OP of this answer in your own post for helping you arrive at the final solution if you do not wish to accept this answer as it stands. $\endgroup$ – Brahadeesh Apr 25 at 11:17

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