ပန်းတိုင်း မွှေးပါစေ
Exercise (9.1) ၏ မေးခွန်း ပုစ္ဆာများကို ဖြေရှင်းရန် ပြဌာန်းစာအုပ်ပါ Theorem (1) မှ Theorem (4) အထိ သီအိုရမ် မှန်ကန်ချက်များကို ဦးစွာလေ့လာထားရမည် ဖြစ်ပါသည်။ Theorem (1) မှ Theorem (4) အထိ ရှင်းလင်းချက်များကို ဒီ post မှာ လေ့လာနိုင်ပါသည်။
Ex 9.1-No (1)
The point $O$ is the centre of the given circle. Find the values of $x$, $y$ and $z$.
(a)
| $\begin{array}{l} x=\angle A D C(\angle \mathrm{s} \text { on arc } A C) \\\\ \therefore x=70^{\circ}\\\\ A B C D \text { is cyclic quadrilateral. }\\\\ 70^{\circ}+\beta=180^{\circ}\\\\ \therefore \beta=180^{\circ}-70^{\circ}=110^{\circ} \\\\ \text { Since } A D / / \mathrm{BC}, \\\\ \theta=\angle C A D \\\\ \therefore \quad \theta=30^{\circ} \\\\ \therefore \quad y\ =180^{\circ}-(\beta+\theta) \\\\ \quad \quad\quad =180^{\circ}-\left(110^{\circ}+30^{\circ}\right) \\\\ \quad\quad\quad =40^{\circ} \end{array}$ |
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| $\begin{array}{l} \text { Since } A C=A D, \angle A C D=x \\\\ \text{In}\ \Delta A C D, x+x=150^{\circ} \\\\ \therefore\ 2 x=150^{\circ} \Rightarrow x=75^{\circ}\\\\ A B C D \text { is a cyclic quadrilateral. }\\\\ \therefore x+\phi=180^{\circ}\\\\ \therefore\ \ \phi\ =180^{\circ}-x \\\\ \quad\quad =180^{\circ}-75^{\circ} \\\\ \quad\quad =105^{\circ} \\\\ \text { Since } A B=B C, \angle B A C=y . \\\\ \text { In } \Delta A B C, 2 y+\phi=180^{\circ} \\\\ \therefore\ 2 y=180^{\circ}-\phi \\\\ \quad\quad\ =180^{\circ}-105^{\circ} \\\\ \quad\ y\ \ =37.5^{\circ} \end{array}$ |
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| $\begin{aligned} \angle B O D &=360^{\circ}-240^{\circ} \\\\ & =120^{\circ} \\\\ x &=\displaystyle\frac{1}{2} \angle B O D \\\\ &=\displaystyle\frac{1}{2}\left(120^{\circ}\right) \\\\ &=60^{\circ} \\\\ \theta &=\displaystyle\frac{1}{2}\left(240^{\circ}\right) \\\\ \therefore\quad\theta &=120^{\circ} \\\\ \therefore\quad y &=360^{\circ}-\left(120^{\circ}+120^{\circ}+55^{\circ}\right) \\\\ &=65^{\circ} \end{aligned}$ |
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In $\Delta A D E$, $\theta=180^{\circ}-\left(30^{\circ}+40^{\circ}\right)=110^{\circ}$ $A C D E$ is a cyclic quadrilateral. $\begin{aligned} \therefore y+ \theta &=180^{\circ} \\\\ \therefore y \quad &=180^{\circ}-\theta \\\\ &=180^{\circ}-110^{\circ} \\\\ &=70^{\circ} \end{aligned}$ In $\Delta A C D, A C =A D$. $\therefore\quad \angle A D C=y$ $A C D E$ is a cyclic quadrilateral. $\begin{aligned} \therefore\quad x+y & =180^{\circ} \\\\ \therefore\quad x \quad &=180^{\circ}-y \\\\ &=180^{\circ}-70^{\circ} \\\\ &=110^{\circ} \end{aligned}$ |
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Since $O A=O B\ \text{( radii )}$ $\therefore\quad \theta=50^{\circ}$ $A B D E$ is a cyclic quadrilateral. $\therefore\quad x+60^{\circ}+\theta=180^{\circ}$ $\begin{aligned} \therefore \quad x & =180^{\circ}-\left(60^{\circ}+\theta\right) \\\\ & =180^{\circ}-\left(60^{\circ}+50^{\circ}\right) \\\\ & =70^{\circ} \end{aligned}$ $C E$ is a diameter. $\therefore \angle C D E=90^{\circ}$ $\begin{aligned} \therefore \quad x+y & =90^{\circ}\\\\ \therefore \quad y & =90^{\circ}-x\\\\ & =90^{\circ}-70^{\circ}\\\\ & =20^{\circ} \end{aligned}$ |
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$A B D E$ is a cyclic quadrilateral. $\therefore y=110^{\circ}$ Since $\text{arc}\ A D=\text{arc}\ B C,$ $A D=B C$ $\therefore\quad A B D E$ is isosceles trapezium. $\therefore\quad x=y=110^{\circ}$ $x+z=180^{\circ}$ $\therefore\quad z=180^{\circ}-x$ $\quad\quad =180^{\circ}-110^{\circ}$ $\quad\quad =70^{\circ}$ |
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Ex 9.1-No (2)
$ABC$ is an acute triangle inscribed in $\odot O$, and $OD$ is the perpendicular drawn $O$ to $BC$. Prove that $\angle BOD = \angle BAC$.
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Ex 9.1-No (3)
In $\odot O$, two chords $A B$ and $C D$ intersect in the circle at $P$. Show that $\angle A P D=\displaystyle\frac{1}{2}(\angle A O D+\angle B O C)$.
 In $\triangle A P C$, $\begin{aligned} \angle A P D & =\angle P C A+\angle P A C \\\\ & =\angle D C A+\angle B A C \\\\ & =\displaystyle\frac{1}{2} \angle A O D+\displaystyle\frac{1}{2} \angle B O C \\\\ & =\displaystyle\frac{1}{2}(\angle A O D+\angle B O C) \end{aligned}$ |
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Ex 9.1-No (4)
Two circles intersect at $M, N$. From $M$, diameters $M A, M B$ are drawn in each circle. If $A, B$ are joined to $N$, prove that $A N B$ is a straight line.
| $\begin{array}{lll} \text{Given} &: & \odot O\ \text{and}\ \odot P\ \text{intersect at}\ M \text{and}\ N.\\\\ & & M A\ \text{and}\ M B\ \text{are diameters of}\ \odot O\ \text{and}\\\\ & & \odot P\ \text{and respectively.}\\\\ \text{To Prove} &: & A N B\ \text{is a straight line.}\\\\ \text{Proof} &: & \text{Join}\ M N.\\\\ & & \text{In}\ \odot O, \angle A N M=90^{\circ} \quad(AM\ \text{is a diameter} )\\\\ & & \text{In}\ \odot P, \angle B N M=90^{\circ} \quad(BM\ \text{is a diameter} )\\\\ & & \therefore\quad \angle A N M+\angle B N M=180^{\circ}\\\\ & & \therefore\quad A N B\ \text{ is a straight line.} \end{array}$ |
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Ex 9.1-No (5)
$OA$ and $O B$ are two radii of a circle meeting at right angles. From $A, B$ two parallel chords $A X, B Y$ are drawn. Prove that $A Y \perp B X$.
| $\begin{array}{lll} \text{Given} &: & O A \perp O B, A X \parallel B Y\\\\ \text{To Prove} &: & A Y \perp B X .\\\\ \text{Proof} &: & \text{Let}\ A Y\ \text{cut}\ B X\ \text{at}\ D \\\\ & & \alpha=\gamma \quad(\angle\mathrm{s}\ \text{stand on arc}\ AB )\\\\ & & \text{But}\ \alpha=\gamma=\displaystyle\frac{1}{2} \angle A O B \\\\ & & \alpha=\gamma=\displaystyle\frac{1}{2}\left(90^{\circ}\right)=45^{\circ}\\\\ & & \text{Since} A X \parallel B Y,\\\\ & & \alpha=\beta \quad (\text{alternating} \angle \mathrm{s})\\\\ & & \therefore\quad \beta=45^{\circ}\\\\ & & \text{Since} \angle A D B=\beta+\gamma,\\\\ & & \angle A D B=90^{\circ}\\\\ & & \therefore\quad A Y \perp B X.\\\\ \end{array}$ |
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Ex 9.1-No (6)
Two circles intersect at $R$ and $S$. Two straight lines $A R B$ and $C S D$ are drawn meeting one circle at $A, C$ and the other at $B, D$. Prove that $A C \parallel B D .$ If $A B \parallel C D$, show that $A B=C D$.
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Join $RS$ $ACSR$ is a cyclic quadrilateral. $\therefore\quad \alpha=\theta$ $BDSR$ is also a cyclic quadrilateral. $\therefore\quad \beta+\theta = 180^{\circ}$ $\therefore\quad \alpha+\beta= 180^{\circ}$ $\therefore\quad A C \parallel B D$ If $A B \parallel C D$, $ABDC$ is a parallelogram. then we can say $A B=C D$ |
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