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In the figure, AB is diameter, PM = MH and PN = NB. If PEB arc = 108 °, calculate "x" enter image description here

Drawing to obtain the right triangle in P: $\Delta APB $. $\hat{A}$ as $\hat{B}$ are angles inscribed on the circumference, they are equal to half the arc you see. Therefore, $\hat{A} = 54^o$ , $\hat{B} = 36^o$ and note also that $\overline {MN}$ t is average base $\Delta HPB $.
If $\hat{B} = 36^o$. $\hat{H} = 90^o$ then $H\hat{P}B = 54^o$

enter image description here

In the right triangle, the middle base is parallel to the base of the original triangle. Therefore, the triangle $\Delta PMN$ is $\hat{M}$ rectangle is $P \hat{N} M = 36^o$ and In addition $R \hat{Q}N$ and $O \hat{Q}M$ are opposed by the vertex. From this, it is easy to conclude that $\Delta OMQ$ $\simeq$ $\Delta NRQ$ Therefore, in the triangle $\Delta OMQ$ we obtain that $ x + 90^o + 36^o = 180^o \Rightarrow x = 54^o $

how to demonstrate the similarity of $\Delta OMQ$ $ and \Delta NRQ$ ?

peta arantes
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    show that $ \angle AOH = \angle ARP$ so $\angle MRN =\angle 180-AOH$ this shows that $OMRN$ is a cyclic quadrilateral and we are done – endgame yourgame Nov 26 '20 at 08:04
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    There are two errors in your text. $\measuredangle PMN$ is equal to $90^0$ and not $36^0$, However, $\measuredangle PNM$ is equal to $36^0$. Furthermore, $\measuredangle RQN$ and $\measuredangle OQN$ are adjacent angles and not opposite angles. But, $\measuredangle RQN$ and $\measuredangle MQO$ are opposite angles. – YNK Nov 27 '20 at 08:54
  • Thank´s ..I've already corrected – peta arantes Nov 27 '20 at 18:15
  • $OR\parallel AP\to x=54°$ – Raffaele Nov 27 '20 at 18:46
  • I need an algebraic demonstration and not just an affirmation – peta arantes Nov 29 '20 at 14:50

1 Answers1

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Angle Alpha

The sketch shown in $\mathrm{Fig.\space 1}$ is an extended version of the diagrams you have included in your problem statement. We denote the sought angle $\measuredangle RQN$ as $\alpha$. Let $F$ be the center of the semicircle $BPA$ and $\measuredangle NRO=\phi$.

As you have mentioned in your incomplete answer, $\measuredangle PAB$ and $\measuredangle ABP$ can be easily determined as $54^o $ and $36^o$ respectively, because $\measuredangle PFB =108^o$. It is also given that $M$ and $N$ are the midpoints of the segments $PH$ and $PB$ respectively. Therefore $MN$ is parallel to $AB$, which makes $\measuredangle MNR = 36^o$.

To solve this seemingly difficult problem we have to bring Ceva’s theorem in to play. That is why we have extended the line $NM$ to intersect $PA$ at $D$. This makes $D$ the midpoint of $PA$. As you see below, solving of the problem now becomes as easy as eating a piece of cake. Consider the triangle $PAN$, in which three cevians $ND$, $PO$, and $AR$ are concurrent at $M$. Applying Ceva’s theorem to $\triangle PAN$, we obtain, $$\frac{PD}{DA}\frac{AO}{ON}\frac{NR}{RP}=1. \tag{1}$$

Since $PD=DA$, equation (1) can be simplified to get, $$\frac{AO}{ON}=\frac{RP}{NR}. \tag{2}$$

Equation (2) implies that $PA$ is parallel to $RO$. Therefore, we have $\phi = \measuredangle NRO = \measuredangle BPA = 90^o$. Using triangle $RQN$, we obtain, $$\alpha = \measuredangle RQN =180^o-\measuredangle QNR\space\space –\space\phi =180^o – 36^o – 90^o = 54^o.$$

YNK
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