In \( \triangle \mathrm{ABC} \), the perpendiculars AM and BN drawn from vertices A and B on their opposite sides meet (when produced) the circumcircle of the triangle at point D and E respectively. Prove that: (a) \( \quad \mathrm{ABMN} \) is a cyclic quadrilateral. (b) \( \quad \operatorname{arc} \mathrm{CD}=\operatorname{arc} C E \)

(a) To prove that \( \mathrm{ABMN} \) is a cyclic quadrilateral, we need to show that opposite angles are supplementary. Since AM and BN are perpendiculars from A and B to the opposite sides of \( \triangle \mathrm{ABC} \), we have: - \( \angle AMB = 90^\circ \) (because AM is perpendicular to BC) - \( \angle BNA = 90^\circ \) (because BN is perpendicular to AC) Now, consider the angles at M and N in quadrilateral \( \mathrm{ABMN} \): - \( \angle AMN \) is the external angle at M, which is supplementary to \( \angle AMB \), so \( \angle AMN = 180^\circ - 90^\circ = 90^\circ \). - \( \angle BNM \) is the external angle at N, which is supplementary to \( \angle BNA \), so \( \angle BNM = 180^\circ - 90^\circ = 90^\circ \). Since \( \angle AMN \) and \( \angle BNM \) are both right angles, \( \mathrm{ABMN} \) is a cyclic quadrilateral because it has two right angles that are opposite each other. (b) To prove that \( \operatorname{arc} \mathrm{CD} = \operatorname{arc} \mathrm{CE} \), we can use the fact that \( \mathrm{ABMN} \) is a cyclic quadrilateral. In a cyclic quadrilateral, opposite angles are supplementary. Since \( \angle AMN \) and \( \angle BNM \) are both right angles, they are supplementary to each other. This means that the angles subtended by the same arc at the center of the circle are equal. Let \( \angle AMD \) and \( \angle BNE \) be the angles subtended by arc \( \mathrm{CD} \) and \( \mathrm{CE} \) at the center of the circle. Since \( \mathrm{ABMN} \) is cyclic, \( \angle AMD \) and \( \angle BNE \) are supplementary, which means \( \angle AMD + \angle BNE = 180^\circ \). Now, since \( \angle AMD \) and \( \angle BNE \) are subtended by the same arc \( \mathrm{CD} \) and \( \mathrm{CE} \), they are equal. Therefore, \( \operatorname{arc} \mathrm{CD} = \operatorname{arc} \mathrm{CE} \).