Question

note: figure not drawn to scale. in the figure shown, $overline{ba}$ is congruent to $overline{de}$. what is the value of $\theta$?

Explanation:

Step1: Identify triangle congruence

Triangles \( \triangle BAE \) and \( \triangle DEA \) are right triangles (since \( \angle BAE = \angle DEA = 90^\circ \)), and \( BA \cong DE \), \( AE \) is common. So \( \triangle BAE \cong \triangle DEA \) (HL congruence). Thus, \( \angle BAE = \angle DEA = 90^\circ \), and \( \angle B = 63^\circ \), so in \( \triangle BAE \), \( \angle BAE = 90^\circ \), \( \angle B = 63^\circ \), so \( \angle BEA = 180^\circ - 90^\circ - 63^\circ = 27^\circ \).

Step2: Find \( \theta \) using linear pair or triangle angles

Since \( \angle BCA \) and \( \angle DCE \) are vertical angles, and in triangle \( \triangle ACE \), but more simply, \( \theta = 90^\circ + 27^\circ = 117^\circ \)? Wait, no. Wait, \( \triangle ABC \) and \( \triangle DEC \): \( BA = DE \), right angles, and \( AC = EC \) (since \( \triangle BAE \cong \triangle DEA \), so \( BE = DA \), and \( AC = EC \) as midpoints? Wait, better: in right triangles \( \triangle BAC \) and \( \triangle DEC \), \( BA = DE \), \( \angle BAC = \angle DEC = 90^\circ - \angle BEA \)? Wait, no. Wait, \( \angle B = 63^\circ \), \( \angle BAC = 90^\circ \), so \( \angle BCA = 180^\circ - 90^\circ - 63^\circ = 27^\circ \). Then \( \theta = 180^\circ - 27^\circ = 153^\circ \)? No, wait, vertical angles: \( \angle BCA \) and \( \angle DCE \) are equal. Wait, no, let's re-examine.

Wait, \( \triangle BAE \) and \( \triangle DEA \) are congruent (HL: right angle, \( BA = DE \), \( AE = AE \)). So \( \angle BEA = \angle DAE \). In \( \triangle BAE \), \( \angle B = 63^\circ \), \( \angle BAE = 90^\circ \), so \( \angle BEA = 27^\circ \). Then \( \angle DAE = 27^\circ \). Now, in triangle \( ACE \), \( \angle CAE = 90^\circ - \angle DAE = 63^\circ \)? No, wait, \( \angle BAE = 90^\circ \), so \( \angle BAC + \angle CAE = 90^\circ \). Wait, maybe better: \( \theta \) is the angle at \( C \), which is vertical to \( \angle BCA \)'s supplement? Wait, no. Let's use the fact that \( \triangle ABC \) and \( \triangle DEC \) are congruent (AAS: \( \angle B = \angle D \) (since triangles \( BAE \) and \( DEA \) are congruent), \( \angle BAC = \angle DEC = 90^\circ \), \( BA = DE \)). So \( \angle BCA = \angle DCE \). In \( \triangle ABC \), \( \angle B = 63^\circ \), \( \angle BAC = 90^\circ \), so \( \angle BCA = 27^\circ \). Then \( \theta = 180^\circ - 27^\circ = 153^\circ \)? No, wait, \( \theta \) is adjacent to \( \angle BCA \). Wait, no, the angle at \( C \) between the two diagonals: \( \theta \) and \( \angle BCA \) are supplementary? Wait, no, \( \angle BCA \) and \( \theta \) are vertical angles? No, \( \angle BCA \) and \( \angle DCE \) are vertical angles, and \( \theta \) is \( 180^\circ - \angle BCA \)? Wait, no, let's calculate the angle at \( C \).

Wait, \( \angle B = 63^\circ \), \( \angle BAC = 90^\circ \), so \( \angle BCA = 27^\circ \). Then \( \theta = 180^\circ - 27^\circ = 153^\circ \)? No, that's not right. Wait, maybe \( \theta = 63^\circ + 90^\circ = 153^\circ \)? Wait, no, let's think again.

Wait, \( BA \cong DE \), \( \angle BAE = \angle DEA = 90^\circ \), \( AE \) is common. So \( \triangle BAE \cong \triangle DEA \) (SAS: \( BA = DE \), \( \angle BAE = \angle DEA \), \( AE = AE \)). Thus, \( \angle AEB = \angle DAE \). In \( \triangle BAE \), \( \angle AEB = 90^\circ - 63^\circ = 27^\circ \), so \( \angle DAE = 27^\circ \). Now, \( \angle BAC = 90^\circ \), so \( \angle BAC = \angle BAE = 90^\circ \), so \( \angle BAC = 90^\circ \), \( \angle DAE = 27^\circ \), so \( \angle CAD = 90^\circ - 27^\circ = 63^\circ \)? No, this is getting confusing. Wait, the key is that \( \theta \) is the sum of \( 90^\circ \) and \( 63^\circ \)? No, wait, \( \theta = 180^\circ - (90^\circ - 63^\circ) = 153^\circ \). Wait, let's use the exterior angle theorem. The angle at \( C \), \( \theta \), is equal to \( \angle B + \angle BAE \)? No, \( \angle B = 63^\circ \), \( \angle BAE = 90^\circ \), so \( \theta = 63^\circ + 90^\circ = 153^\circ \). Yes, that makes sense. Because in the triangle, the exterior angle is equal to the sum of the two non-adjacent interior angles. So \( \theta = 63^\circ + 90^\circ = 153^\circ \).

Step1: Calculate angle in right triangle

In \( \triangle BAE \), \( \angle B = 63^\circ \), \( \angle BAE = 90^\circ \), so \( \angle AEB = 180^\circ - 90^\circ - 63^\circ = 27^\circ \).

Step2: Find \( \theta \) using linear pair or exterior angle

\( \theta \) is supplementary to \( \angle AEB \) in the vertical angle? No, wait, \( \theta = 180^\circ - 27^\circ = 153^\circ \)? Wait, no, \( \theta \) is actually the sum of \( 90^\circ \) and \( 63^\circ \) because of the congruent triangles and right angles. Wait, let's correct:

Since \( \triangle BAE \cong \triangle DEA \), \( \angle B = \angle D = 63^\circ \), \( \angle BAE = \angle DEA = 90^\circ \). Then, looking at the intersection at \( C \), \( \theta \) is equal to \( \angle B + \angle BAE = 63^\circ + 90^\circ = 153^\circ \). Wait, no, that's not the exterior angle. Wait, the angle at \( C \): \( \angle BCA = 27^\circ \) (from \( \triangle ABC \), \( 90 - 63 = 27 \)), then \( \theta = 180 - 27 = 153 \). Yes, that's correct. So \( \theta = 153^\circ \).

Answer:

\( \boxed{153} \)