Prove that triangle ADC is congruent to triangle ABC (use two different methods). 2. Show that AC bisects DCB.

From the diagram markings, we are given that $\angle CAD = \angle CAB$ (marked as $A_1$ and $A_2$) and $\angle ACD = \angle ACB$ (marked as $C_1$ and $C_2$).

• $\angle CAD = \angle CAB$ (Given from diagram markings)
• $AC = AC$ (Common side to both triangles)
• $\angle ACD = \angle ACB$ (Given from diagram markings) Therefore, by the Angle-Side-Angle (ASA) congruence criterion, $\triangle ADC \equiv \triangle ABC$.

Since AC bisects $\angle DAB$ (because $\angle CAD = \angle CAB$) and AC bisects $\angle DCB$ (because $\angle ACD = \angle ACB$), the quadrilateral ABCD is a kite with AC as its axis of symmetry. In a kite, the two pairs of adjacent sides are equal.

• $AD = AB$ (Property of a kite, as AC bisects $\angle DAB$)
• $CD = CB$ (Property of a kite, as AC bisects $\angle DCB$)
• $AC = AC$ (Common side to both triangles) Therefore, by the Side-Side-Side (SSS) congruence criterion, $\triangle ADC \equiv \triangle ABC$.

From Step 1 or Step 2, we have proven that $\triangle ADC \equiv \triangle ABC$.
By the property of congruent triangles, corresponding parts are equal (CPCTC).
Therefore, the corresponding angles $\angle ACD$ and $\angle ACB$ must be equal.
Since $\angle ACD = \angle ACB$, the line segment AC divides $\angle DCB$ into two equal angles.
Thus, AC bisects $\angle DCB$.

The question labeled "c)" provides information about an isosceles trapezium and a rectangle but does not pose a question to be solved.

$\boxed{\text{1. } \triangle ADC \equiv \triangle ABC \text{ (by ASA and SSS)}}$
$\boxed{\text{2. AC bisects } \angle DCB}$