Question

9.
m∠abc = m∠bcd = 90°
m∠cda = m∠dab = 90°

Explanation:

Step1: Identify the figure type

The figure has four right angles, so it's a rectangle (or square - type quadrilateral with all angles 90°). Triangles \( \triangle ABD \) and \( \triangle BCD \) are congruent? Wait, actually, since all angles of the quadrilateral are 90°, it's a rectangle, so \( AB \parallel CD \) and \( AD \parallel BC \), and \( BD \) is a diagonal? Wait, no, looking at the angles at \( D \): \( \angle CDB = 35^\circ \), and \( \angle CDA = 90^\circ \), so \( \angle ADB=\theta \), and \( \theta + 35^\circ=90^\circ \) (since \( \angle CDA = 90^\circ \) is the sum of \( \angle CDB \) and \( \angle ADB \)).

Step2: Solve for \( \theta \)

Since \( \angle CDA = 90^\circ \), and \( \angle CDA=\angle CDB+\angle ADB \), with \( \angle CDB = 35^\circ \) and \( \angle ADB=\theta \), we have:
\( \theta+ 35^\circ=90^\circ \)
Subtract \( 35^\circ \) from both sides:
\( \theta=90^\circ - 35^\circ=55^\circ \)

Also, for the angles at \( B \), since the quadrilateral is a rectangle, \( \triangle ABD \) and \( \triangle BCD \) are congruent right - triangles. So \( \angle ABD=\theta = 55^\circ \) (because in right - triangle \( \triangle ABD \), \( \angle ADB=\theta \) and \( \angle DAB = 90^\circ \), so \( \angle ABD = 90^\circ-\theta \)? Wait, no, earlier we found \( \theta = 55^\circ \) from \( \angle CDA \). Wait, let's re - check:

In \( \triangle BCD \), \( \angle BCD = 90^\circ \), \( \angle CDB = 35^\circ \), so \( \angle CBD=90^\circ - 35^\circ = 55^\circ \). Since the quadrilateral is a rectangle, \( AB = CD \), \( AD = BC \), and \( BD \) is common, so \( \triangle ABD\cong\triangle CDB \) (by SAS, as \( AB = CD \), \( \angle DAB=\angle BCD = 90^\circ \), \( AD = BC \)). Therefore, \( \angle ABD=\angle CDB = 35^\circ \)? Wait, no, I think I made a mistake. Wait, the quadrilateral has all angles 90°, so it's a rectangle, so \( \angle ADB=\angle CBD \) (alternate interior angles, since \( AD\parallel BC \)). Wait, \( \angle CDB = 35^\circ \), \( \angle BCD = 90^\circ \), so \( \angle CBD=55^\circ \), so \( \angle ADB = 55^\circ \) (alternate interior angles, \( AD\parallel BC \), transversal \( BD \)). And in \( \triangle ABD \), \( \angle DAB = 90^\circ \), so \( \angle ABD=90^\circ - 55^\circ = 35^\circ \)? Wait, no, let's start over.

The quadrilateral \( ABCD \) has \( \angle DAB=\angle ABC=\angle BCD=\angle CDA = 90^\circ \), so it's a rectangle. So \( AD\parallel BC \) and \( AB\parallel CD \).

In \( \triangle BCD \): \( \angle BCD = 90^\circ \), \( \angle CDB = 35^\circ \), so \( \angle CBD=90 - 35=55^\circ \).

Since \( AD\parallel BC \), \( \angle ADB=\angle CBD \) (alternate interior angles, transversal \( BD \)), so \( \theta=\angle ADB = 55^\circ \).

For the angles at \( B \): \( \angle ABC = 90^\circ=\angle ABD+\angle CBD \), \( \angle CBD = 55^\circ \), so \( \angle ABD = 90 - 55 = 35^\circ \)? Wait, no, earlier when we looked at \( \angle CDA \): \( \angle CDA=90^\circ=\angle CDB+\angle ADB \), \( \angle CDB = 35^\circ \), so \( \angle ADB=\theta = 90 - 35 = 55^\circ \). Then in \( \triangle ABD \), \( \angle DAB = 90^\circ \), so \( \angle ABD=90 - 55 = 35^\circ \). So \( \theta = 55^\circ \), \( b = 35^\circ \), \( c = 55^\circ \) (since \( \angle ABC = 90^\circ=b + c \), and from congruent triangles or alternate interior angles, \( b=\angle CDB = 35^\circ \), \( c=\angle ADB = 55^\circ \)).

Answer:

If we are solving for \( \theta \), \( \theta = 55^\circ \); for \( b \), \( b = 35^\circ \); for \( c \), \( c = 55^\circ \) (assuming we need to find these angles. The most probable angle to find is \( \theta \), so \( \theta=\boldsymbol{55^\circ} \))