Question

Given: start overline, A, D, end overline, is perpendicular to, start overline, D, B, end overline, comma
AD
⊥
DB
, start overline, B, C, end overline, is perpendicular to, start overline, C, A, end overline
BC
⊥
CA
and start overline, A, D, end overline, \cong, start overline, C, B, end overline, .
AD
≅
CB
.

Prove: triangle, A, D, B, \cong, triangle, B, C, A△ADB≅△BCA.
In the table below, the first step of the proof is completed for you. To complete each following step, first use the dropdown menu in the "Statement" column to select the format of a statement. Once you select a format, a statement will appear with input boxes and/or dropdowns to complete, and a dropdown menu will appear in the reason column. Once you complete these inputs and dropdowns, navigate back to the first column and press the "try" button. If your statement and reason are valid, a new row will appear below. This process will continue until you have filled in a statement and reason for what you were asked to prove, and then the "try" button will cause the answer area to appear with a submit button. The diagram associated with this problem is located below the proof table, and is available as a tactile printout. There may be a clarifying note below the diagram.
Step Statement Reason
1
start overline, A, D, end overline, is perpendicular to, start overline, D, B, end overline
AD
⊥
DB

start overline, B, C, end overline, is perpendicular to, start overline, C, A, end overline
BC
⊥
CA

start overline, A, D, end overline, \cong, start overline, C, B, end overline
AD
≅
CB

Given
\cong≅
✕
A
B
C
D
E
An image is shown. To download tactile, press SHIFT-CONTROL-U for UEB or SHIFT-CONTROL-E for Nemeth.
Note: start overline, A, C, end overline
AC
and start overline, D, B, end overline
DB
are segments.

Answer 1

To complete the proof and show that triangle \( ADB \cong BCA \), we will fill in the table step by step.

### Step 1
- **Statement:** \( \overline{AD} \perp \overline{DB} \)
- **Reason:** Given

### Step 2
- **Statement:** \( \overline{BC} \perp \overline{CA} \)
- **Reason:** Given

### Step 3
- **Statement:** \( \overline{AD} \cong \overline{CB} \)
- **Reason:** Given

### Step 4
- **Statement:** \( \angle ADB \cong \angle BCA \)
- **Reason:** Both are right angles (since \( AD \perp DB \) and \( BC \perp CA \))

### Step 5
- **Statement:** \( \triangle ADB \cong \triangle BCA \)
- **Reason:** Right Angle-Hypotenuse-Side (RHS) Congruence Theorem

Now that all the steps are completed, ensure each step logically follows from the previous ones and matches the provided information to conclude that \( \triangle ADB \cong \triangle BCA \).

You can press the "try" button now to check if these entries are valid, and then submit your proof!

Answer 2

so would this be an hl

Answer 3

Yes, the proof can be considered as HL (Hypotenuse-Leg) congruence.

The components of the HL criterion you've used include:

1. **Right Angles**: Both triangles \( ADB \) and \( BCA \) contain right angles:
- \( \angle ADB \) and \( \angle BCA \) are both right angles because \( \overline{AD} \perp \overline{DB} \) and \( \overline{BC} \perp \overline{CA} \).

2. **Hypotenuse**: The segments \( \overline{AD} \) and \( \overline{CB} \) are given to be congruent:
- \( \overline{AD} \cong \overline{CB} \).

3. **Leg**: The segment \( \overline{DB} \) in triangle \( ADB \) and \( \overline{CA} \) in triangle \( BCA \):
- Since they are each one leg of the respective right triangles and need to be shown congruent.

Using this information, we can conclude that triangle \( ADB \) is congruent to triangle \( BCA \) by HL. The right angles, one pair of congruent legs (the hypotenuses), and the right angle condition satisfy the HL theorem criteria.

So yes, you would classify the proof as using the Hypotenuse-Leg criterion for right triangles.