Given:
Segment AD is perpendicular to segment BC.
D is the midpoint of segment BC.
To prove:
[tex]\Delta BDA\cong \Delta CDA[/tex]
Solution:
The two column proof is shown below:
Statements Reasons
1. [tex]\overline{AD}\perp \overline{BC}[/tex], D is the midpoint of [tex]\overline{BC}[/tex] 1. Given
2. [tex]\overline{BD}\cong \overline{CD}[/tex] 2. Definition of midpoint
3. [tex]m\angle ADB=90^\circ \text{ and }m\angle ADC=90^\circ[/tex] 3. Definition of perpendicular lines
4. [tex]m\angle ADB=m\angle ADC[/tex] 4. Substitution
5. [tex]m\angle ADB\cong m\angle ADC[/tex] 5. Definition of congruence
6. [tex]\overline{AD}\cong \overline{AD}[/tex] 6. Reflexive property
7. [tex]\Delta BDA\cong \Delta CDA[/tex] 7. SAS congruence postulate
Hence proved.
