2. prove: the diagonals of a rhombus bisect the angles to which they are drawn. note: you may not use any other properties of rhombi/parallelograms as reasons. given: ab = bc = cd = da prove: ∠1 ≅ ∠2 and ∠3 ≅ ∠4

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# Explanation:
## Step1: Analyze Triangle $ \triangle ABD $
In rhombus $ ABCD $, $ AB = AD $ (given $ AB = BC = CD = DA $). Consider diagonal $ BD $. In $ \triangle ABD $, $ AB = AD $, so $ \triangle ABD $ is isosceles with $ \angle 1 $ and $ \angle 2 $ as base angles? Wait, no, let's use $ \triangle ABC $ and $ \triangle ADC $? Wait, better to use $ \triangle ABD $ and $ \triangle CBD $. Wait, given $ AB = AD $ (since $ AB = DA $) and $ AB = BC $, $ AD = DC $. Let's take diagonal $ AC $ first for $ \angle 1 \cong \angle 2 $.

In $ \triangle ABC $ and $ \triangle ADC $:
- $ AB = AD $ (given)
- $ BC = DC $ (given)
- $ AC = AC $ (common side)

So by SSS (Side - Side - Side) congruence criterion, $ \triangle ABC \cong \triangle ADC $. Wait, no, for $ \angle 1 $ and $ \angle 2 $, let's take $ \triangle ABD $? Wait, no, $ \angle 1 $ is $ \angle DAC $, $ \angle 2 $ is $ \angle BAC $. Let's consider $ \triangle ADB $ and $ \triangle CDB $? Wait, maybe better to take $ \triangle ABC $ and $ \triangle ADC $ is not right. Wait, let's start with diagonal $ AC $ to prove $ \angle 1 \cong \angle 2 $.

In $ \triangle ADB $: $ AB = AD $ (given, since $ AB = DA $). Wait, no, $ AB = AD $, so $ \triangle ABD $ is isosceles with $ AB = AD $. Then diagonal $ AC $ intersects $ BD $ at $ E $. Wait, maybe the problem is to prove that diagonal $ AC $ bisects $ \angle DAB $ (so $ \angle 1 \cong \angle 2 $) and diagonal $ BD $ bisects $ \angle ABC $ (so $ \angle 3 \cong \angle 4 $).

Let's prove $ \angle 1 \cong \angle 2 $ first. Consider $ \triangle ABC $ and $ \triangle ADC $:
- $ AB = AD $ (given: $ AB = DA $)
- $ BC = DC $ (given: $ BC = CD $)
- $ AC = AC $ (common side)

By SSS congruence, $ \triangle ABC \cong \triangle ADC $. Wait, no, $ \triangle ABC $ has sides $ AB, BC, AC $, $ \triangle ADC $ has sides $ AD, DC, AC $. Since $ AB = AD $, $ BC = DC $, $ AC = AC $, so $ \triangle ABC \cong \triangle ADC $. Then corresponding angles: $ \angle BAC \cong \angle DAC $, which is $ \angle 2 \cong \angle 1 $, so $ \angle 1 \cong \angle 2 $.

Now for $ \angle 3 \cong \angle 4 $. Consider $ \triangle ABD $ and $ \triangle CBD $:
- $ AB = CB $ (given)
- $ AD = CD $ (given)
- $ BD = BD $ (common side)

By SSS congruence, $ \triangle ABD \cong \triangle CBD $. Then corresponding angles: $ \angle ABD \cong \angle CBD $, which is $ \angle 3 \cong \angle 4 $.

Wait, let's re - structure:

## Step1: Prove $ \angle 1 \cong \angle 2 $ (diagonal $ AC $ bisects $ \angle DAB $)
Consider $ \triangle ABC $ and $ \triangle ADC $:
- $ AB = AD $ (given: $ AB = DA $)
- $ BC = DC $ (given: $ BC = CD $)
- $ AC = AC $ (common side)

By SSS congruence, $ \triangle ABC \cong \triangle ADC $.
Since corresponding parts of congruent triangles are congruent (CPCTC), $ \angle BAC \cong \angle DAC $, i.e., $ \angle 2 \cong \angle 1 $ or $ \angle 1 \cong \angle 2 $.

## Step2: Prove $ \angle 3 \cong \angle 4 $ (diagonal $ BD $ bisects $ \angle ABC $)
Consider $ \triangle ABD $ and $ \triangle CBD $:
- $ AB = CB $ (given: $ AB = BC $)
- $ AD = CD $ (given: $ AD = CD $)
- $ BD = BD $ (common side)

By SSS congruence, $ \triangle ABD \cong \triangle CBD $.
Since corresponding parts of congruent triangles are congruent (CPCTC), $ \angle ABD \cong \angle CBD $, i.e., $ \angle 3 \cong \angle 4 $.

# Answer:
To prove that the diagonals of a rhombus bisect the angles to which they are drawn (i.e., $ \angle 1\cong\angle 2 $ and $ \angle 3\cong\angle 4 $):

### Part 1: Prove $ \angle 1\cong\angle 2 $
1. **Identify Triangles and Sides**: Consider $ \triangle ABC $ and $ \triangle ADC $.
   - Given $ AB = AD $ (since $ AB = DA $), $ BC = DC $ (since $ BC = CD $), and $ AC = AC $ (common side).
2. **SSS Congruence**: By the Side - Side - Side (SSS) congruence criterion, $ \triangle ABC\cong\triangle ADC $.
3. **CPCTC (Corresponding Parts of Congruent Triangles are Congruent)**: Since $ \triangle ABC\cong\triangle ADC $, their corresponding angles $ \angle BAC $ (which is $ \angle 2 $) and $ \angle DAC $ (which is $ \angle 1 $) are congruent. Thus, $ \angle 1\cong\angle 2 $.

### Part 2: Prove $ \angle 3\cong\angle 4 $
1. **Identify Triangles and Sides**: Consider $ \triangle ABD $ and $ \triangle CBD $.
   - Given $ AB = CB $ (since $ AB = BC $), $ AD = CD $ (since $ AD = CD $), and $ BD = BD $ (common side).
2. **SSS Congruence**: By the SSS congruence criterion, $ \triangle ABD\cong\triangle CBD $.
3. **CPCTC**: Since $ \triangle ABD\cong\triangle CBD $, their corresponding angles $ \angle ABD $ (which is $ \angle 3 $) and $ \angle CBD $ (which is $ \angle 4 $) are congruent. Thus, $ \angle 3\cong\angle 4 $.

Therefore, the diagonals of a rhombus bisect the angles to which they are drawn.

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze Triangle \( \triangle ABD \)

Step1: Prove \( \angle 1 \cong \angle 2 \) (diagonal \( AC \) bisects \( \angle DAB \))

Step2: Prove \( \angle 3 \cong \angle 4 \) (diagonal \( BD \) bisects \( \angle ABC \))

Answer:

Part 1: Prove \( \angle 1\cong\angle 2 \)

Part 2: Prove \( \angle 3\cong\angle 4 \)