inscribed triangle problem

Jun 2010
34
2
'ABC is an acute-angled triangle inscribed in a circle and P, Q, R are the midpoints of the minor arcs BC, CA, AB respectively.
Prove that AP is perpendicular to QR.'
I know that the lines from the midpoints to the centre of the circle are perpendicular bisectors of the sides of the triangle
but I'm struggling to know how to use this.
 
Jun 2013
1,152
615
Lebanon
AP and QR meet at M

measure of angle PMQ =

\(\displaystyle \frac{\text{arc} \text{PQ} + \text{arc} \text{AR} }{2}=\frac{\text{arc} \text{PC} + \text{arc} \text{CQ} + \text{arc} \text{AR}}{2}=90 \text{degrees}\)
 
Jun 2010
34
2
Idea - thank you for responding but I don't understand your answer. When you write arcPQ, do you mean the angle subtended at the centre of the circle by arc PQ?
 

Plato

MHF Helper
Aug 2006
22,507
8,664
Idea - thank you for responding but I don't understand your answer. When you write arcPQ, do you mean the angle subtended at the centre of the circle by arc PQ?
$m(\angle ABC) = \frac{1}{2}m\left( {arcAC} \right)$ the inscribed angle theorem.

Note that $m\left( {arcAQ}= \right)m\left( {arcQC} \right)=\frac{1}{2}m\left( {arcAC} \right).$

Use the intersecting chord theorem to show that $m(\angle AGQ) = \dfrac{\pi }{2}$, where $\left\{ G \right\} = \overline {AP} \cap \overline {RQ} $
 
Jun 2013
1,152
615
Lebanon
let's talk about angles

central angle is twice inscribed angle so \(\displaystyle ROP=2RQP\) where \(\displaystyle O\) is the center of the circle

similarly \(\displaystyle AOQ=2APQ\)

\(\displaystyle 4*RQP + 4*APQ = 2*ROP + 2*AOQ = 360\) because this is once around the circle

therefore \(\displaystyle RQP + APQ = 90\)

\(\displaystyle MQP + MPQ = 90\) in triangle \(\displaystyle MPQ\)

angle \(\displaystyle M=90\)
 

Plato

MHF Helper
Aug 2006
22,507
8,664
let's talk about angles

central angle is twice inscribed angle so \(\displaystyle ROP=2RQP\) where \(\displaystyle O\) is the center of the circle

similarly \(\displaystyle AOQ=2APQ\)

\(\displaystyle 4*RQP + 4*APQ = 2*ROP + 2*AOQ = 360\) because this is once around the circle

therefore \(\displaystyle RQP + APQ = 90\)

\(\displaystyle MQP + MPQ = 90\) in triangle \(\displaystyle MPQ\)

angle \(\displaystyle \color{red}M=90\)
Is that the number ninety or some ninety mystery units like degrees?
 
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