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Question
complete the proof by the following
23)
given:
prove:
24)
given:
prove:
To solve the problem of proving the congruence of triangles, we will use the Side - Side - Side (SSS) or Side - Angle - Side (SAS) congruence criteria. Let's analyze the two given triangle - related problems one by one.
Problem 1 (First Triangle Diagram)
Step 1: Identify the given information
From the diagram and the given data, we know that \(AB\parallel DC\) (given), \(AB = DC\) (given), and \(AC\) is a common side. Also, since \(AB\parallel DC\), \(\angle BAC=\angle DCA\) (alternate interior angles).
Step 2: Choose the congruence criterion
We have two sides and the included angle equal. \(AB = DC\), \(\angle BAC=\angle DCA\), and \(AC = AC\). So, by the SAS (Side - Angle - Side) congruence criterion, \(\triangle ABC\cong\triangle CDA\).
Problem 2 (Second Triangle Diagram: Prove \(\triangle ABD\cong\triangle CBD\))
Step 1: Identify the given information
We are given that \(AB = CB\), \(AD = CD\), and \(BD\) is a common side.
Step 2: Choose the congruence criterion
We have three sides equal: \(AB = CB\), \(AD = CD\), and \(BD=BD\). So, by the SSS (Side - Side - Side) congruence criterion, \(\triangle ABD\cong\triangle CBD\).
For the first proof ( \(\triangle ABC\cong\triangle CDA\)):
| Statement | Reason |
|---|---|
| 2. \(\angle BAC=\angle DCA\) | Alternate interior angles (since \(AB\parallel DC\)) |
| 3. \(AB = DC\) | Given |
| 4. \(AC = AC\) | Common side |
| 5. \(\triangle ABC\cong\triangle CDA\) | SAS (Steps 2, 3, 4) |
For the second proof ( \(\triangle ABD\cong\triangle CBD\)):
| Statement | Reason |
|---|---|
| 2. \(AD = CD\) | Given |
| 3. \(BD = BD\) | Common side |
| 4. \(\triangle ABD\cong\triangle CBD\) | SSS (Steps 1, 2, 3) |
If we consider the first proof:
Step 1: Identify alternate interior angles
Since \(AB\parallel DC\), \(\angle BAC=\angle DCA\) (alternate interior angles theorem).
Step 2: List equal sides
We know \(AB = DC\) (given) and \(AC = AC\) (common side).
Step 3: Apply SAS criterion
By SAS (\(AB = DC\), \(\angle BAC=\angle DCA\), \(AC = AC\)), \(\triangle ABC\cong\triangle CDA\).
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\(\triangle ABC\cong\triangle CDA\) (by SAS) and \(\triangle ABD\cong\triangle CBD\) (by SSS)