NCERT Solutions · Class 9 Mathematics I’m Up and Down, and Round and Round

Look at the arc that each side cuts off.The idea. Take one side, say \(\displaystyle AB \). The line \(\displaystyle AB \) splits the plane in two, and the other two vertices \(\displaystyle C \) and \(\displaystyle D \) are on one of those sides — the inside of the quadrilateral is on that side too. The centre \(\displaystyle O \) lies on the same side of the chord \(\displaystyle AB \) as the major arc. So:\[O \text{ is inside } ABCD \iff \text{every side cuts off an arc smaller than a semicircle.} \]Turning that into something you can measure. The arc \(\displaystyle AB \) that does not contain \(\displaystyle C \) and \(\displaystyle D \) is measured by the angle it subtends at \(\displaystyle C \) (or at \(\displaystyle D \)): \(\displaystyle \angle ACB = \tfrac{1}{2}\,\text{arc } AB \). Therefore, for each side, look at the angle it subtends at one of the two opposite vertices: So one honest method is: check all four such angles. All acute \(\displaystyle \Rightarrow \) centre inside.The best way — you only ever need one measurement. The four arcs add up to \(\displaystyle 360^{\circ} \), so at most one of them can exceed \(\displaystyle 180^{\circ} \). Moreover, the side cutting off that big arc is always the longest side. Here is why. Suppose arc \(\displaystyle AB = a > 180^{\circ} \). Then the other three arcs satisfy \(\displaystyle b + c + d = 360^{\circ} - a < 180^{\circ} \). Now the chord \(\displaystyle AB \) also spans the minor arc of measure \(\displaystyle 360^{\circ} - a = b+c+d \), and each of \(\displaystyle b, c, d \) is smaller than that. Among minor arcs (arcs under \(\displaystyle 180^{\circ} \)) of one circle, a longer arc has a longer chord. Hence \(\displaystyle AB \) is longer than \(\displaystyle BC \), \(\displaystyle CD \) and \(\displaystyle DA \).So the quick test is:1. Find the longest side of the quadrilateral. 2. Measure the angle it subtends at either of the two remaining vertices (the two answers are equal). 3. Acute \(\displaystyle \Rightarrow \) the centre is inside; exactly \(\displaystyle 90^{\circ} \Rightarrow \) the centre is on that side (its midpoint); obtuse \(\displaystyle \Rightarrow \) the centre is outside, beyond that side.Sanity checks. For a rectangle every side subtends an acute angle at the far vertices, and indeed the centre is inside (it is where the diagonals cross). For a thin quadrilateral whose four vertices are bunched together on a small piece of the circle, the long "closing" side subtends a fat obtuse angle, and the centre is far outside.An equivalent way of saying it, if you prefer triangles: a diagonal cuts the quadrilateral into two triangles inscribed in the same circle, and the circumcentre of a triangle lies inside it exactly when the triangle is acute-angled. The centre lies inside the quadrilateral precisely when at least one of those two triangles is not obtuse.Test the longest side: if it subtends an acute angle at the opposite vertices the centre is inside, if a right angle the centre lies on that side, and if an obtuse angle the centre is outside. Since at most one side can cut off a major arc, and that side must be the longest one, this single check settles it.

Equal chords are equidistant from the centre, then RHS congruence.Let the circle have centre \(\displaystyle O \) and radius \(\displaystyle r \). Let the two chords be \(\displaystyle AB \) and \(\displaystyle CD \) with \(\displaystyle AB = CD \), crossing at a point \(\displaystyle P \) inside the circle. Let \(\displaystyle M \) be the midpoint of \(\displaystyle AB \) and \(\displaystyle N \) the midpoint of \(\displaystyle CD \), and join \(\displaystyle OM \), \(\displaystyle ON \), \(\displaystyle OP \).Step $\displaystyle 1$: \(\displaystyle OM \perp AB \) and \(\displaystyle ON \perp CD \). The line from the centre to the midpoint of a chord is perpendicular to that chord (this is the perpendicular-bisector result of this chapter).Step $\displaystyle 2$: the two chords are the same distance from the centre. Write \(\displaystyle h \) for the common half-length, so \[h = AM = MB = CN = ND = \tfrac{1}{2}AB = \tfrac{1}{2}CD . \] In the right triangles \(\displaystyle OMA \) and \(\displaystyle ONC \), Pythagoras gives \[OM^{2} = r^{2} - h^{2} = ON^{2}, \qquad\text{so}\qquad OM = ON. \]Step $\displaystyle 3$: \(\displaystyle P \) is the same distance from both midpoints. Compare \(\displaystyle \triangle OMP \) and \(\displaystyle \triangle ONP \): By RHS congruence, \(\displaystyle \triangle OMP \cong \triangle ONP \), so \[MP = NP. \] Call this common distance \(\displaystyle d \).Step $\displaystyle 4$: read off the four pieces. On the chord \(\displaystyle AB \), the point \(\displaystyle P \) sits at distance \(\displaystyle d \) from the midpoint \(\displaystyle M \), so the two pieces of \(\displaystyle AB \) are \[h + d \quad\text{and}\quad h - d. \] On the chord \(\displaystyle CD \), the point \(\displaystyle P \) sits at the same distance \(\displaystyle d \) from the midpoint \(\displaystyle N \), so the two pieces of \(\displaystyle CD \) are also \[h + d \quad\text{and}\quad h - d. \]So the pieces of one chord match the pieces of the other: one of \(\displaystyle PC \), \(\displaystyle PD \) equals \(\displaystyle PA \), and the other equals \(\displaystyle PB \). (Which name goes with which just depends on how you labelled the endpoints.)Special case: if \(\displaystyle d = 0 \) then \(\displaystyle P \) is the midpoint of both chords, so both chords are bisected at \(\displaystyle P \) — which happens exactly when both are diameters and \(\displaystyle P = O \).An alternative, if you have met the intersecting-chords relation \(\displaystyle PA \cdot PB = PC \cdot PD \): here also \(\displaystyle PA + PB = AB = CD = PC + PD \). Two positive numbers are completely determined by their sum and their product, so the pair \(\displaystyle \{PA, PB\} \) must be the same pair as \(\displaystyle \{PC, PD\} \).Hence if two intersecting chords are equal in length, the segments of one are equal to the corresponding segments of the other.

Build the right triangle first; its circumcircle is the circle you want.First, what circle are we after? Let \(\displaystyle AB \) be the required chord, \(\displaystyle AB = 6 \) cm, and let \(\displaystyle O \) be the centre with \(\displaystyle OM \perp AB \), \(\displaystyle OM = 3 \) cm. Then \(\displaystyle AM = 3 \) cm, so triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \) with both legs equal to $\displaystyle 3$ cm, and \[OA^{2} = 3^{2} + 3^{2} = 18, \qquad OA = 3\sqrt{2} \approx 4.2 \text{ cm}. \] Also \(\displaystyle \angle AOM = 45^{\circ} \), so \(\displaystyle \angle AOB = 90^{\circ} \): the chord subtends a right angle at the centre.The hint's construction. A chord subtending \(\displaystyle 90^{\circ} \) at the centre is a leg of an inscribed right isosceles triangle. So take the triangle with a right angle and two equal sides of $\displaystyle 6$ cm, and draw its circumcircle:1. Draw a right angle at a point \(\displaystyle A \). 2. On one arm mark \(\displaystyle B \) with \(\displaystyle AB = 6 \) cm; on the other arm mark \(\displaystyle D \) with \(\displaystyle AD = 6 \) cm. 3. Join \(\displaystyle BD \). (It is the hypotenuse, \(\displaystyle BD = \sqrt{6^{2}+6^{2}} = 6\sqrt{2} \approx 8.5 \) cm.) 4. Mark \(\displaystyle O \), the midpoint of \(\displaystyle BD \). Because the angle in a semicircle is a right angle, the hypotenuse of a right triangle is a diameter of its circumcircle, so the circumcentre is the midpoint of \(\displaystyle BD \). 5. With centre \(\displaystyle O \) and radius \(\displaystyle OA = OB = OD = 3\sqrt{2} \approx 4.2 \) cm, draw the circle. It passes through \(\displaystyle A \), \(\displaystyle B \) and \(\displaystyle D \).Then \(\displaystyle AB \) is a chord of this circle of length $\displaystyle 6$ cm, and its distance from \(\displaystyle O \) is $\displaystyle 3$ cm — exactly what was asked. (So is \(\displaystyle AD \).)Verification with coordinates. Put \(\displaystyle A = (0,0) \), \(\displaystyle B = (6,0) \), \(\displaystyle D = (0,6) \). Then \(\displaystyle O \), the midpoint of \(\displaystyle BD \), is \(\displaystyle (3,3) \). Distances: \(\displaystyle OA = OB = OD = \sqrt{3^{2}+3^{2}} = 3\sqrt{2} \), so all three points really are on one circle centred at \(\displaystyle O \). The chord \(\displaystyle AB \) lies along the \(\displaystyle x \)-axis, so its distance from \(\displaystyle O = (3,3) \) is \(\displaystyle 3 \) cm, and its length is \(\displaystyle 6 \) cm. Correct.A shorter route, if you do not want a triangle at all: draw \(\displaystyle AB = 6 \) cm, construct its perpendicular bisector, mark \(\displaystyle O \) on that bisector $\displaystyle 3$ cm from \(\displaystyle AB \), and draw the circle with centre \(\displaystyle O \) and radius \(\displaystyle OA \). This works because the centre of any circle must lie on the perpendicular bisector of each of its chords.Draw a right isosceles triangle with legs $\displaystyle 6$ cm and take its circumcircle: centre at the midpoint of the hypotenuse, radius \(\displaystyle 3\sqrt{2} \approx 4.2 \) cm. Each $\displaystyle 6$ cm leg is then a chord $\displaystyle 3$ cm from the centre.

Combine "opposite angles of a parallelogram are equal" with "opposite angles of a cyclic quadrilateral are supplementary".Suppose a parallelogram \(\displaystyle ABCD \) is inscribed in a circle. Two facts now apply to the same pair of angles.Because \(\displaystyle ABCD \) is a parallelogram, opposite angles are equal: \[\angle A = \angle C. \]Because \(\displaystyle ABCD \) is cyclic, opposite angles are supplementary: \[\angle A + \angle C = 180^{\circ}. \]Substituting the first into the second, \[\angle A + \angle A = 180^{\circ} \quad\Longrightarrow\quad 2\angle A = 180^{\circ} \quad\Longrightarrow\quad \angle A = 90^{\circ}, \] and hence \(\displaystyle \angle C = 90^{\circ} \) too. The same argument on the other pair gives \(\displaystyle \angle B = \angle D = 90^{\circ} \). (Or simply: the angles of a quadrilateral add to \(\displaystyle 360^{\circ} \), so the remaining two share \(\displaystyle 180^{\circ} \) equally.)A parallelogram with all its angles right angles is by definition a rectangle. So the only parallelogram that can be inscribed in a circle is a rectangle.And a rectangle really can be. This matters — otherwise the statement would be empty. Let \(\displaystyle ABCD \) be a rectangle and \(\displaystyle P \) the point where its diagonals meet. The diagonals of a rectangle are equal and bisect each other, so \[PA = PB = PC = PD = \tfrac{1}{2}(\text{diagonal}). \] All four vertices are the same distance from \(\displaystyle P \), so the circle with centre \(\displaystyle P \) and that radius passes through all four — the rectangle is cyclic.Every cyclic parallelogram is a rectangle, and every rectangle is cyclic; so the rectangle is the only parallelogram that can be inscribed in a circle.

A right angle inscribed in a circle stands on a diameter.Let the rectangle \(\displaystyle ABCD \) be inscribed in a circle with centre \(\displaystyle O \), and let \(\displaystyle P \) be the point where its diagonals \(\displaystyle AC \) and \(\displaystyle BD \) cross. We must show \(\displaystyle P = O \).Step $\displaystyle 1$: each diagonal is a diameter. Look at \(\displaystyle \angle ABC \). It is an angle of a rectangle, so \(\displaystyle \angle ABC = 90^{\circ} \); and it is an inscribed angle standing on the arc \(\displaystyle AC \) not containing \(\displaystyle B \). An inscribed angle is half its arc, so \[\text{arc } AC = 2 \times 90^{\circ} = 180^{\circ}. \] An arc of \(\displaystyle 180^{\circ} \) is a semicircle, so its chord \(\displaystyle AC \) is a diameter, and therefore \(\displaystyle AC \) passes through \(\displaystyle O \).The identical argument with \(\displaystyle \angle BCD = 90^{\circ} \) shows that \(\displaystyle BD \) is a diameter too, so \(\displaystyle BD \) also passes through \(\displaystyle O \).Step $\displaystyle 2$: finish. Both diagonals pass through \(\displaystyle O \). Two distinct lines meet in at most one point, and the diagonals of a rectangle are distinct lines meeting at \(\displaystyle P \). Hence \[P = O. \]A second proof, using distances. In a rectangle the diagonals are equal and bisect each other, so \[PA = PB = PC = PD = \tfrac{1}{2}(\text{diagonal}). \] So \(\displaystyle P \) is equidistant from all four vertices, i.e. \(\displaystyle P \) is the centre of a circle through \(\displaystyle A, B, C, D \). But three non-collinear points lie on exactly one circle, so that circle is the given circle and \(\displaystyle P \) is its centre.The diagonals of an inscribed rectangle are diameters, so their point of intersection is the centre of the circle.

Every such midpoint is the same distance from the centre.Let the circle have centre \(\displaystyle O \) and radius \(\displaystyle r \), and let all the chords have the same fixed length \(\displaystyle \ell \) (with \(\displaystyle \ell \le 2r \), since no chord can be longer than a diameter).One direction: every midpoint is at a fixed distance from \(\displaystyle O \). Take any chord \(\displaystyle AB \) of length \(\displaystyle \ell \), with midpoint \(\displaystyle M \). The line from the centre to the midpoint of a chord is perpendicular to the chord, so \(\displaystyle \triangle OMA \) is right angled at \(\displaystyle M \), with \(\displaystyle AM = \tfrac{\ell}{2} \) and \(\displaystyle OA = r \). Pythagoras gives \[OM^{2} = r^{2} - \left(\frac{\ell}{2}\right)^{2}, \qquad OM = \sqrt{\,r^{2} - \frac{\ell^{2}}{4}\,}. \] Nothing on the right depends on which chord of length \(\displaystyle \ell \) we chose. So every one of these midpoints lies at this one fixed distance \(\displaystyle d = \sqrt{r^{2} - \ell^{2}/4} \) from \(\displaystyle O \) — that is, on a circle with centre \(\displaystyle O \) and radius \(\displaystyle d \).The other direction: the whole of that circle is used. Take any point \(\displaystyle M \) with \(\displaystyle OM = d \). Draw the line through \(\displaystyle M \) perpendicular to \(\displaystyle OM \); it cuts the circle at two points \(\displaystyle A \) and \(\displaystyle B \). Since \(\displaystyle OM \perp AB \), \(\displaystyle M \) is the midpoint of \(\displaystyle AB \), and \[AM^{2} = r^{2} - d^{2} = r^{2} - \left(r^{2} - \frac{\ell^{2}}{4}\right) = \frac{\ell^{2}}{4}, \] so \(\displaystyle AB = 2\,AM = \ell \). So \(\displaystyle M \) really is the midpoint of a chord of length \(\displaystyle \ell \). Nothing on that circle is left out.Example: in a circle of radius $\displaystyle 10$ units, the midpoints of all chords of length $\displaystyle 12$ units form a circle of radius \(\displaystyle \sqrt{100-36} = 8 \) units around the same centre.The degenerate case: if \(\displaystyle \ell = 2r \) the chords are diameters, they all have the same midpoint \(\displaystyle O \), and the "circle" of midpoints shrinks to the single point \(\displaystyle O \) (radius \(\displaystyle 0 \)) — which the formula gives correctly.The midpoints form a circle with the same centre as the given circle (a concentric circle), of radius \(\displaystyle \sqrt{\,r^{2} - \ell^{2}/4\,} \).

SSS congruence on the two triangles the radii make.We are given a circle with centre \(\displaystyle O \) and two chords from the same point \(\displaystyle A \), with \(\displaystyle AB = AC \). Join \(\displaystyle OA \), \(\displaystyle OB \), \(\displaystyle OC \).Compare \(\displaystyle \triangle OAB \) and \(\displaystyle \triangle OAC \): So \(\displaystyle \triangle OAB \cong \triangle OAC \) by SSS. Corresponding angles of congruent triangles are equal, so \[\angle OAB = \angle OAC. \]But \(\displaystyle \angle OAB \) and \(\displaystyle \angle OAC \) are the two parts into which the ray \(\displaystyle AO \) divides \(\displaystyle \angle BAC \). Since these two parts are equal, \(\displaystyle AO \) is the bisector of \(\displaystyle \angle BAC \). Hence \(\displaystyle O \) lies on that bisector, which is the statement we wanted.Another way to see it. Both \(\displaystyle A \) and \(\displaystyle O \) are equidistant from \(\displaystyle B \) and \(\displaystyle C \) — \(\displaystyle A \) because \(\displaystyle AB = AC \), and \(\displaystyle O \) because \(\displaystyle OB = OC \) are radii. Two distinct points equidistant from \(\displaystyle B \) and \(\displaystyle C \) determine the perpendicular bisector of \(\displaystyle BC \), so the line \(\displaystyle AO \) is the perpendicular bisector of \(\displaystyle BC \). In the isosceles triangle \(\displaystyle ABC \) (with \(\displaystyle AB = AC \)), the perpendicular bisector of the base is also the bisector of the apex angle \(\displaystyle \angle BAC \). So \(\displaystyle O \) lies on the bisector of \(\displaystyle \angle BAC \).A caution worth noticing: this says the centre is on the bisector, not that it is inside the triangle. If \(\displaystyle \angle BAC \) is obtuse the centre still lies on the bisecting ray, just beyond \(\displaystyle BC \).Because \(\displaystyle OB = OC \) and \(\displaystyle AB = AC \), the triangles \(\displaystyle OAB \) and \(\displaystyle OAC \) are congruent, so \(\displaystyle AO \) splits \(\displaystyle \angle BAC \) into two equal parts — the centre lies on the angle bisector.

Two right triangles sharing the same radius.Let \(\displaystyle O \) be the centre and \(\displaystyle r \) the radius. Let \(\displaystyle AB = 24 \) cm and \(\displaystyle CD = 10 \) cm be the two parallel chords, both on the same side of \(\displaystyle O \). Drop the perpendicular from \(\displaystyle O \); because the chords are parallel, one single perpendicular line meets both, at \(\displaystyle M \) on \(\displaystyle AB \) and \(\displaystyle N \) on \(\displaystyle CD \), and these are the midpoints of the chords.Write \(\displaystyle OM = p \) and \(\displaystyle ON = q \). Half-lengths are \(\displaystyle AM = 12 \) cm and \(\displaystyle CN = 5 \) cm, so Pythagoras in the two right triangles \(\displaystyle OMA \) and \(\displaystyle ONC \) gives \[r^{2} = p^{2} + 12^{2} = p^{2} + 144, \tag{1} \] \[r^{2} = q^{2} + 5^{2} = q^{2} + 25. \tag{2} \]Setting up the $\displaystyle 7$ cm. Both chords are on the same side of \(\displaystyle O \), so \(\displaystyle M \) and \(\displaystyle N \) are on the same side and the gap between the chords is the difference of the distances: \[|q - p| = 7. \] The longer chord is the nearer one to the centre, so \(\displaystyle p < q \) and \[q = p + 7. \]Solving. Put \(\displaystyle q = p+7 \) into ($\displaystyle 2$) and set it equal to ($\displaystyle 1$): \[p^{2} + 144 = (p+7)^{2} + 25 = p^{2} + 14p + 49 + 25, \] \[144 = 14p + 74, \] \[14p = 70, \qquad p = 5. \] Then \(\displaystyle q = 5 + 7 = 12 \), and from ($\displaystyle 1$) \[r^{2} = 5^{2} + 144 = 25 + 144 = 169, \qquad r = 13 \text{ cm}. \]Check the other possibility, to be safe. If instead we had assumed the $\displaystyle 24$ cm chord was the farther one, \(\displaystyle p = q + 7 \), then \(\displaystyle (q+7)^{2} + 144 = q^{2} + 25 \) gives \(\displaystyle 14q + 49 + 144 = 25 \), i.e. \(\displaystyle 14q = -168 \), a negative distance. Impossible — so the solution above is the only one.Check the answer directly. With \(\displaystyle r = 13 \): the chord $\displaystyle 5$ cm from the centre has half-length \(\displaystyle \sqrt{169-25} = 12 \), so length $\displaystyle 24$ cm; the chord $\displaystyle 12$ cm from the centre has half-length \(\displaystyle \sqrt{169-144}=5 \), so length $\displaystyle 10$ cm; and the two are \(\displaystyle 12 - 5 = 7 \) cm apart on the same side. All three conditions hold.The radius of the circle is $\displaystyle 13$ cm.

Cut the hexagon into six triangles at the centre.Let the circle have centre \(\displaystyle O\) and radius \(\displaystyle r\), and let the regular hexagon be \(\displaystyle ABCDEF\) with all six vertices on the circle. Join \(\displaystyle O\) to every vertex. This cuts the hexagon into $\displaystyle 6$ triangles, and every one of the joining segments is a radius: \[OA = OB = OC = OD = OE = OF = r. \]Step $\displaystyle 1$: each angle at the centre is \(\displaystyle 60^\circ\). The hexagon is regular, so its six sides are equal; together with the equal radii, the six triangles are congruent by SSS. Hence the six angles they make at \(\displaystyle O\) are equal. Those six angles fill up one complete turn about \(\displaystyle O\), so \[\text{each angle at } O = \frac{360^\circ}{6} = 60^\circ. \]Step $\displaystyle 2$: each triangle is equilateral, so the side equals \(\displaystyle r\). Take \(\displaystyle \triangle OAB\). Since \(\displaystyle OA = OB\), it is isosceles, so the two base angles are equal: \[\angle OAB = \angle OBA = \frac{180^\circ - 60^\circ}{2} = 60^\circ. \] All three angles are \(\displaystyle 60^\circ\), so \(\displaystyle \triangle OAB\) is equilateral and \[AB = OA = r. \]Step $\displaystyle 3$: the distance of a side from the centre. Drop \(\displaystyle OM \perp AB\), with \(\displaystyle M\) on \(\displaystyle AB\). The perpendicular from the centre to a chord bisects the chord (the same thing happens here because \(\displaystyle \triangle OAB\) is isosceles), so \[AM = \frac{AB}{2} = \frac{r}{2}. \] Now use Pythagoras in the right triangle \(\displaystyle OMA\): \[OM^2 = OA^2 - AM^2 = r^2 - \frac{r^2}{4} = \frac{3r^2}{4}, \qquad OM = \frac{\sqrt{3}}{2}\,r. \]A quick sanity check. The perimeter of the hexagon is \(\displaystyle 6r\), and the circumference of the circle is \(\displaystyle 2\pi r \approx 6.28r\). The hexagon sits just inside the circle, so its perimeter should be a little less than the circumference — and \(\displaystyle 6r < 6.28r\). Also \(\displaystyle OM = 0.866r\) is less than \(\displaystyle r\), as it must be, since a chord lies nearer the centre than the circle does.Each side of the hexagon has length \(\displaystyle r\), and each side is at a distance \(\displaystyle \dfrac{\sqrt{3}}{2}r \approx 0.87r\) from the centre.

Angles in the same segment.(Note that here \(\displaystyle O\) is one of the four vertices of the quadrilateral, not the centre of the circle.)The four points lie on the circle in the order \(\displaystyle M, N, O, P\). Draw the diagonal \(\displaystyle MP\). Now look at the two angles asked about: \(\displaystyle \angle \mathrm{MNP}\) is the angle that the same* chord \(\displaystyle MP\) subtends at the point \(\displaystyle N\).The chord \(\displaystyle MP\) splits the circle into two arcs. Going round the circle in order \(\displaystyle M \to N \to O \to P\), both \(\displaystyle N\) and \(\displaystyle O\) lie on the same one of these two arcs — that is, \(\displaystyle N\) and \(\displaystyle O\) are points of the same segment standing on \(\displaystyle MP\).Angles in the same segment are equal, because each of them is half of the one central angle standing on the arc \(\displaystyle MP\). So \[\angle \mathrm{MOP} = \angle \mathrm{MNP}. \]Where the diameter comes in. The equality above did not actually need \(\displaystyle MN\) to be a diameter. What the diameter tells you is the size of these angles. Since \(\displaystyle MN\) is a diameter, the angle in the semicircle at \(\displaystyle P\) is a right angle: \[\angle \mathrm{MPN} = 90^\circ . \] So in the right triangle \(\displaystyle MPN\), \[\angle \mathrm{MNP} = 90^\circ - \angle \mathrm{PMN}, \] and therefore \[\angle \mathrm{MOP} = \angle \mathrm{MNP} = 90^\circ - \angle \mathrm{PMN}. \] (For the same reason \(\displaystyle \angle \mathrm{MON} = 90^\circ\) as well, since \(\displaystyle O\) also sees the diameter \(\displaystyle MN\).)Checked numerically: placing \(\displaystyle M\) and \(\displaystyle N\) at the two ends of a diameter and trying many positions of \(\displaystyle O\) and \(\displaystyle P\) on the arc, the three quantities \(\displaystyle \angle \mathrm{MOP}\), \(\displaystyle \angle \mathrm{MNP}\) and \(\displaystyle 90^\circ - \angle \mathrm{PMN}\) agreed every time (for example \(\displaystyle 13.47^\circ\), \(\displaystyle 43.62^\circ\), \(\displaystyle 81.98^\circ\) in three different positions).\(\displaystyle \angle \mathrm{MOP} = \angle \mathrm{MNP}\): they are equal, being angles in the same segment standing on the chord \(\displaystyle MP\); and because \(\displaystyle MN\) is a diameter each of them equals \(\displaystyle 90^\circ - \angle \mathrm{PMN}\).