Question

1. as shown in the figure, in rectangle abcd, diagonals ac and bd intersect at point o. draw ae⊥bd through a such that the perpendicular - foot is located at e. if m∠ocd = 36°, then m∠eao = ______. 4. which of the following conditions can be used to prove that parallelogram abcd is a rectangle? a. cd = ab b. m∠a = m∠c c. ad = ab d. ac = bd

Explanation:

Step1: Recall rectangle properties

In rectangle \(ABCD\), \(OA = OC\), \(OB=OD\), and \(AC = BD\), so \(OA=OB\). Also, \(AB\parallel CD\), then \(\angle BAC=\angle OCD = 36^{\circ}\).

Step2: Find \(\angle ABO\)

Since \(OA = OB\) in rectangle, \(\angle ABO=\angle BAC = 36^{\circ}\).

Step3: Calculate \(\angle AOB\)

In \(\triangle AOB\), using the angle - sum property of a triangle (\(\angle AOB+ \angle ABO+\angle BAC=180^{\circ}\)), \(\angle AOB = 180^{\circ}-36^{\circ}-36^{\circ}=108^{\circ}\).

Step4: Calculate \(\angle EAO\)

Since \(AE\perp BD\), in right - triangle \(AEO\), \(\angle EAO=90^{\circ}-\angle AOE\). And \(\angle AOE = \angle AOB = 108^{\circ}\), so \(\angle EAO=90^{\circ}- 54^{\circ}=18^{\circ}\).

Step5: Answer the second question

1. Recall the properties of a rectangle:

• In a parallelogram, if the diagonals are equal, then the parallelogram is a rectangle.
• Option A: \(CD = AB\) is a property of all parallelograms (opposite sides are equal), not a specific property to prove a parallelogram is a rectangle.
• Option B: \(m\angle A=m\angle C\) is a property of all parallelograms (opposite angles are equal), not a specific property to prove a parallelogram is a rectangle.
• Option C: \(AD = AB\) means the parallelogram is a rhombus (adjacent sides are equal), not a rectangle.
• Option D: \(AC = BD\) is the condition that can be used to prove that parallelogram \(ABCD\) is a rectangle.

Answer:

1. \(18\)
2. D. \(AC = BD\)