1. **Problem statement:** Given a kite formed by an equilateral triangle BCD and an isosceles triangle ABD with \(\angle ABC = 90^\circ\), find \(\angle BAD\) and other labelled angles with reasons. 2. **Step 1: Analyze the kite and triangles** - Since BCD is equilateral, all its angles are \(60^\circ\). - Since ABD is isosceles with AB = AD (kite property), \(\angle ABD = \angle ADB\). - Given \(\angle ABC = 90^\circ\), and BCD is equilateral, \(\angle CBD = 60^\circ\). - Therefore, \(\angle ABD = 90^\circ - 60^\circ = 30^\circ\). 3. **Step 2: Find \(\angle BAD\)** - Triangle ABD has angles \(\angle ABD = 30^\circ\), \(\angle ADB = 30^\circ\), so \(\angle BAD = 180^\circ - 30^\circ - 30^\circ = 120^\circ\). 4. **Step 3: Angles in square ABEF** - All angles in square ABEF are \(90^\circ\). - \(\angle ABC = 90^\circ\) and \(\angle FED = 90^\circ\) since ABC and FED are straight lines. 5. **Step 4: Angles in parallelograms and rhombuses** - Given angles 62° and 110° in parallelograms, opposite angles are equal. - Rhombuses ABGH, BCFG, and CDEF have equal sides and opposite equal angles. 6. **Step 5: Triangle EGH angles** - Given angles 6°, 62°, and 9°, sum to \(6 + 62 + 9 = 77^\circ\), so the remaining angle is \(180^\circ - 77^\circ = 103^\circ\). 7. **Step 6: Prove \(c = e\) given \(b + d = 180^\circ\)** - Sum of angles in quadrilateral: \(a + b + c + d = 360^\circ\). - Substitute \(b + d = 180^\circ\) gives \(a + c + 180^\circ = 360^\circ\). - Simplify: \(a + c = 180^\circ\). - Since \(a + c\) and \(a + b\) are supplementary (angles on a line), and \(b + d = 180^\circ\), it follows that \(c = e\). 8. **Step 7: Prove \(a + b + c + d = 360^\circ\)** - By definition, the sum of interior angles of any quadrilateral is \(360^\circ\). **Final answers:** - \(\angle BAD = 120^\circ\) - Angles in square ABEF are all \(90^\circ\) - \(c = e\) proven by angle sum and supplementary angles - Sum of angles in quadrilateral is \(360^\circ\)