Question

what is m∠bce?
m∠bce = \boxed{^\circ}

Explanation:

Step1: Identify congruent triangles

Triangles \( CDE \) and \( CBE \) are right triangles ( \( \angle D = \angle B = 90^\circ \) ), and \( DE = BE \) (marked congruent), \( CE \) is common. So \( \triangle CDE \cong \triangle CBE \) (HL congruence).

Step2: Find \( \angle DCE \)

Given \( \angle DCE = 64^\circ \), since \( \triangle CDE \cong \triangle CBE \), \( \angle BCE=\angle DCE \)? Wait, no, wait. Wait, \( \angle DCB \) is a straight line? Wait, no, \( \angle D = 90^\circ \), \( \angle B = 90^\circ \), \( DE = BE \), \( CE \) is angle bisector? Wait, actually, in right triangles \( CDE \) and \( CBE \), \( DE = BE \), \( CE = CE \), so they are congruent. Then \( \angle DCE=\angle BCE \)? Wait, no, the angle at \( C \): \( \angle DCB \) is a straight angle? Wait, no, \( \angle D = 90^\circ \), \( \angle B = 90^\circ \), so \( CD \) and \( CB \) are both perpendicular to \( DE \) and \( BE \) respectively. Wait, maybe \( \angle DCE = 64^\circ \), and we need to find \( \angle BCE \). Wait, actually, since \( \triangle CDE \cong \triangle CBE \), then \( \angle DCE=\angle BCE \)? No, wait, maybe \( \angle DCB \) is 180 - 64? No, wait, let's think again. The right angles: \( \angle D = 90^\circ \), \( \angle B = 90^\circ \), \( DE = BE \), \( CE \) is common. So by HL, \( \triangle CDE \cong \triangle CBE \). Therefore, \( \angle DCE = \angle BCE \)? Wait, no, the angle given is \( 64^\circ \) at \( C \) between \( CD \) and \( CE \). Wait, maybe \( \angle BCE = 64^\circ \)? No, wait, no. Wait, \( \angle DCB \) is a straight line? No, \( CD \) and \( CB \) are both perpendicular to \( DE \) and \( BE \), so \( CD \parallel CB \)? No, \( \angle D = 90^\circ \), \( \angle B = 90^\circ \), so \( CD \) and \( CB \) are both perpendicular to \( CE \) related? Wait, maybe the correct approach is: since \( \triangle CDE \cong \triangle CBE \), then \( \angle DCE = \angle BCE \). Wait, but the angle between \( CD \) and \( CE \) is \( 64^\circ \), so \( \angle BCE = 64^\circ \)? No, wait, no. Wait, maybe \( \angle DCB \) is 180 degrees, but \( \angle D = 90^\circ \), \( \angle B = 90^\circ \), so \( CD \) and \( CB \) are perpendicular to \( DE \) and \( BE \), so \( DE \parallel BE \)? No, \( DE \) and \( BE \) meet at \( E \). Wait, maybe the triangles are congruent, so \( \angle BCE = \angle DCE = 64^\circ \)? Wait, no, that can't be. Wait, maybe I made a mistake. Wait, the right angle at \( D \) and \( B \), \( DE = BE \), \( CE \) is common, so \( \triangle CDE \cong \triangle CBE \) (HL). Therefore, \( \angle DCE = \angle BCE \). Wait, but the angle between \( CD \) and \( CE \) is \( 64^\circ \), so \( \angle BCE = 64^\circ \)? No, wait, maybe the angle at \( C \) in \( \triangle CDE \) is \( 64^\circ \), so \( \angle DCE = 64^\circ \), and since the triangles are congruent, \( \angle BCE = 64^\circ \). Wait, but that seems too easy. Wait, maybe the correct answer is \( 64^\circ \)? Wait, no, wait, maybe \( \angle DCB \) is 180 - 64 - 64? No, that doesn't make sense. Wait, let's check again. The diagram: \( CD \perp DE \), \( CB \perp BE \), \( DE = BE \), \( CE \) is the hypotenuse. So by HL, \( \triangle CDE \cong \triangle CBE \). Therefore, corresponding angles \( \angle DCE = \angle BCE \). So if \( \angle DCE = 64^\circ \), then \( \angle BCE = 64^\circ \). Wait, but maybe the angle given is \( \angle DCE = 64^\circ \), so \( \angle BCE = 64^\circ \). Wait, but maybe I'm wrong. Wait, no, the congruence gives that the angles at \( C \) are equal. So \( m\angle BCE = 64^\circ \)? Wait, no, wait, maybe the angle is \( 26^\circ \)? Wait, no, let's think again. Wait, \( \angle D = 90^\circ \), so in \( \triangle CDE \), \( \angle DEC = 90^\circ - 64^\circ = 26^\circ \). Then, since \( \triangle CDE \cong \triangle CBE \), \( \angle BEC = 26^\circ \), so in \( \triangle CBE \), \( \angle BCE = 90^\circ - 26^\circ = 64^\circ \). Oh, right! So that's correct. So \( m\angle BCE = 64^\circ \)? Wait, no, wait, \( \angle D = 90^\circ \), so \( \angle DCE + \angle DEC = 90^\circ \), so \( \angle DEC = 90 - 64 = 26^\circ \). Then, since \( \triangle CDE \cong \triangle CBE \), \( \angle BEC = \angle DEC = 26^\circ \), and \( \angle B = 90^\circ \), so \( \angle BCE + \angle BEC = 90^\circ \), so \( \angle BCE = 90 - 26 = 64^\circ \). Yes, that's correct. So the measure of \( \angle BCE \) is \( 64^\circ \).

Answer:

\( 64 \)