angle outside the circle

[logistic_guy]

\(\displaystyle \bold{Given}\): \(\displaystyle \text{\large$\odot$}\text{Q}, \overline{PR} \perp \overline{ST}\).

\(\displaystyle \bold{Prove}: \overline{PS} \cong \overline{PT}\).

 

[The Highlander]

Facile!

PR bisects ST, therefore, ΔPRS ≅ ΔPST, so PS is the same length as PT.

 

[Aion]

Facile!

PR bisects ST, therefore, ΔPRS ≅ ΔPST, so PS is the same length as PT.

I think you meant [imath]\triangle PRS \cong \triangle PRT[/imath] (by side–angle–side) instead of [imath]\triangle PRS \cong \triangle PST[/imath].

 

[The Highlander]

I think you meant [imath]\triangle PRS \cong \triangle PRT[/imath] (by side–angle–side) instead of [imath]\triangle PRS \cong \triangle PST[/imath].

You're absolutely right.

I copied: "ΔABC ≅ ΔXYZ" in (from my 'stock' text file) and then, in haste, replaced the wrong letters; my bad.

Thank you for spotting it and posting the correct version.

 

[logistic_guy]

Facile!

PR bisects ST, therefore, ΔPRS ≅ ΔPST, so PS is the same length as PT.

Your answer is very long!



You will have to wait to see my very short answer.

 

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{step \ 1}}\)

\(\displaystyle \text{\large$\odot$}\text{Q}, \overline{PR} \perp \overline{ST} \rightarrow \textcolor{grey}{\large\bold{given}}\)

 

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{step \ 2}}\)

\(\displaystyle \overline{PR} \text{ bisects } \overline{ST}\)

\(\displaystyle \bold{Reason}\): In a circle, if a radius (or a line containing the radius) is perpendicular to a chord, then it bisects the chord.

 

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{step \ 3}}\)

\(\displaystyle \overline{PR} \text{ is the perpendicular bisector of } \overline{ST}\)

\(\displaystyle \bold{Reason}\): Follows from step \(\displaystyle 1\) and step \(\displaystyle 2\).

 

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{step \ 4}}\)

\(\displaystyle \overline{PS} \cong \overline{PT}\)

\(\displaystyle \bold{Reason}\): Any point on the perpendicular bisector of a segment is equidistant from the segment’s endpoints.

 

[The Highlander]

Your answer is very long!



You will have to wait to see my very short answer.

So where is your "very short answer"?

(And can you prove that: "Any point on the perpendicular bisector of a segment is equidistant from the segment’s endpoints."?)