A pair of a djacent sides of a rectangle are in the perimeter of the rectangle. find the lengths of the sides and hence, the per \( A B C D \) is a rhombus whose diagonals intersect at \( O \). Show that \( \triangle A O B \cong \triangle C O D \). The diagonals of a rectangle \( A B C D \) intersect at \( O \). If \( \angle B O C=70^{\circ} \). Find \( \angle O D \). \( A B C \) and \( A D C \) are two equilateral triangles on a common base \( A C \) (Fig. 11.30 ). Find the angles of the resulting

Let's break down the problem step by step. ### Part 1: Proving \( \triangle AOB \cong \triangle COD \) 1. **Understanding the properties of a rhombus:** - A rhombus has all sides equal. - The diagonals of a rhombus bisect each other at right angles. 2. **Identifying the triangles:** - In rhombus \( ABCD \), the diagonals \( AC \) and \( BD \) intersect at point \( O \). - We need to show that \( \triangle AOB \cong \triangle COD \). 3. **Using congruence criteria:** - **Side-Side-Side (SSS) Congruence:** - \( AO = CO \) (since diagonals bisect each other) - \( BO = DO \) (since diagonals bisect each other) - \( AB = CD \) (all sides of a rhombus are equal) - Therefore, by SSS, \( \triangle AOB \cong \triangle COD \). ### Part 2: Finding \( \angle O D \) 1. **Given:** - \( \angle BOC = 70^\circ \) - Since \( O \) is the intersection of the diagonals, \( \angle AOB + \angle BOC + \angle COD + \angle AOD = 360^\circ \). 2. **Using properties of the rectangle:** - In a rectangle, the diagonals bisect each other and are equal. - Therefore, \( \angle AOB = \angle COD \) and \( \angle BOC = \angle AOD \). 3. **Calculating \( \angle AOD \):** - Let \( \angle AOB = x \) and \( \angle COD = x \). - Then, \( 2x + 70^\circ + 70^\circ = 360^\circ \). - Simplifying gives \( 2x + 140^\circ = 360^\circ \). - Thus, \( 2x = 220^\circ \) and \( x = 110^\circ \). 4. **Finding \( \angle O D \):** - Since \( \angle AOD = \angle BOC = 70^\circ \), we have \( \angle O D = 70^\circ \). ### Part 3: Finding angles in triangles \( ABC \) and \( ADC \) 1. **Understanding the equilateral triangles:** - Both triangles \( ABC \) and \( ADC \) are equilateral, meaning all angles are \( 60^\circ \). 2. **Finding the angles of the resulting figure:** - The angles at point \( A \) in both triangles are \( \angle CAB = 60^\circ \) and \( \angle DAB = 60^\circ \). - Therefore, the angle at point \( A \) in the resulting figure is \( \angle CAB + \angle DAB = 60^\circ + 60^\circ = 120^\circ \). ### Summary of Results: - \( \triangle AOB \cong \triangle COD \) by SSS. - \( \angle O D = 70^\circ \). - The angle at point \( A \) in the resulting figure is \( 120^\circ \).