Question
Find the remaining angles in the following quadrilaterals.
✓
Answer
Here, AEIO is a rhombus (all sides equal).
In $\triangle E A O, A E=A O$, then the angles opposite them are equal.
$\begin{array}{l}\therefore \angle AOE=\angle AEO=20^{\circ} \\\angle AEO=\angle IEO=20^{\circ} \ldots \ldots \ldots \ldots .\end{array}$
(The diagonals of a rhombus bisect its angles)
Also, $\angle AOE =\angle IOE =20^{\circ}$ $\qquad$ (The diagonals of a rhombus bisect its angles)
$\begin{array}{l}\angle E=2 \times \angle A E O=2 \times 20^{\circ}=40^{\circ} \\\angle E=\angle O=40^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) } \\\angle E+\angle A=180^{\circ} \ldots \ldots \ldots . . \text { (The sum of adjacent angles of a rhombus is } 180^{\circ} \text { ) } \\40^{\circ}+\angle A=180^{\circ} \\\angle A=180^{\circ}-40^{\circ} \\\angle A=140^{\circ} \\\angle A=\angle I=140^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) }\end{array}$
In $\triangle E A O, A E=A O$, then the angles opposite them are equal.
$\begin{array}{l}\therefore \angle AOE=\angle AEO=20^{\circ} \\\angle AEO=\angle IEO=20^{\circ} \ldots \ldots \ldots \ldots .\end{array}$
(The diagonals of a rhombus bisect its angles)
Also, $\angle AOE =\angle IOE =20^{\circ}$ $\qquad$ (The diagonals of a rhombus bisect its angles)
$\begin{array}{l}\angle E=2 \times \angle A E O=2 \times 20^{\circ}=40^{\circ} \\\angle E=\angle O=40^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) } \\\angle E+\angle A=180^{\circ} \ldots \ldots \ldots . . \text { (The sum of adjacent angles of a rhombus is } 180^{\circ} \text { ) } \\40^{\circ}+\angle A=180^{\circ} \\\angle A=180^{\circ}-40^{\circ} \\\angle A=140^{\circ} \\\angle A=\angle I=140^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) }\end{array}$
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