Question

Write a proof in two-column form.

Given: $\overleftrightarrow{\mathit{\text{DC}}}\mathit{\perp }\overleftrightarrow{\mathit{\text{BD}}}$; $\angle \mathbf{1}\cong \angle \mathbf{2}$.

Prove:$\overleftrightarrow{\mathit{\text{BA}}}\mathit{\perp }\overleftrightarrow{\mathit{\text{BD}}}$

Short answer

The two-column proof is :

Statement	Reason
$\angle 1=90°$	$\overleftrightarrow{\mathit{\text{DC}}}\mathit{\perp }\overleftrightarrow{\mathit{\text{BD}}}$
$\angle 2=90°$	$\angle 1=90°$and $\angle 1\cong \angle 2$
$\overleftrightarrow{\mathit{\text{BA}}}\mathit{\perp }\overleftrightarrow{\mathit{\text{BD}}}$	$\angle 2=90°$

Step-by-step solution

Step 1. Observe the diagram.

Here, $\overleftrightarrow{DC}\perp \overleftrightarrow{BD}$; $\angle 1\cong \angle 2$.

Step 2. Show the two-column proof.

Statement	Reason
$‎\angle 1=90°$	$\overleftrightarrow{DC}\perp \overleftrightarrow{BD}$
$‎\angle 2=90°$	$\angle 1=90°$and$\angle 1\cong \angle 2$
$\overleftrightarrow{BA}\perp \overleftrightarrow{BD}$	$\angle 2=90°$

Step 3. State the conclusion.

Therefore, $\overleftrightarrow{DC}\perp \overleftrightarrow{BD}$is proved using the information $\overleftrightarrow{DC}\perp \overleftrightarrow{BD}$; $\angle 1\cong \angle 2$ .