Questions: Given: AC ⊥ BD and BD bisects AC. Prove: ΔABD ≅ ΔCBD. Step Statement Reason 1 AC ⊥ BD Given try BD bisects AC Note: the segment AC is a straight segment.

Given: AC ⊥ BD and BD bisects AC. Prove: ΔABD ≅ ΔCBD. Step Statement Reason 1 AC ⊥ BD Given try BD bisects AC Note: the segment AC is a straight segment.
Transcript text: Given: $\overline{A C} \perp \overline{B D}$ and $\overline{B D}$ bisects $\overline{A C}$. Prove: $\triangle A B D \cong \triangle C B D$. \begin{tabular}{|c|c|l|} \hline Step & Statement & Reason \\ \hline 1 & $\overline{A C} \perp \overline{B D}$ & Given \\ \hline try & $\overline{B D}$ bisects $\overline{A C}$ & \\ \hline \end{tabular} Note: the segment $A C$ is a straight segment.
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Solution

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Solution Steps

Step 1: BD bisects AC

Given that BD bisects AC, we know that AD = CD. This is the definition of a bisector.

Step 2: AC ⊥ BD

Given that AC ⊥ BD, we know that ∠BDA and ∠BDC are right angles. Therefore, ∠BDA ≅ ∠BDC because all right angles are congruent.

Step 3: BD is a common side

BD is a side in both triangles ABD and CBD. Thus, BD ≅ BD by the reflexive property.

Final Answer

Triangle ABD is congruent to triangle CBD by SAS (Side-Angle-Side) congruence. We have two congruent sides (AD≅CD and BD≅BD) and a congruent angle between them (∠BDA ≅ ∠BDC).

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