Given:
In [tex]\Delta ABC, \overline{CA}=\overline{CB}[/tex].
[tex]\overline{CD}[/tex] is an altitude drawn from C to [tex]\overline{AB}[/tex].
To prove:
[tex]\overline{CD}[/tex] bisects [tex]\overline{AB}[/tex].
Proof:
In [tex]\Delta ABC[/tex], [tex]\overline{CD}[/tex] is an altitude drawn from C to [tex]\overline{AB}[/tex].
It means, [tex]\Delta ACD\text{ and }\Delta BCD[/tex] are right angle triangles.
In [tex]\Delta ACD\text{ and }\Delta BCD[/tex],
Hypotenuse : [tex]\overline{CA}=\overline{CB}[/tex] [Given]
Leg : [tex]\overline{CD}=\overline{CD}[/tex] [Common]
If the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then by HL postulate both triangles are congruent.
[tex]\Delta ACD\cong \Delta BCD[/tex] [HL postulate]
[tex]\overline{AD}\cong \overline{BD}[/tex] [CPCTC]
[tex]\overline{AD}=\overline{BD}[/tex]
It means, point D is the midpoint of [tex]\overline{AB}[/tex].
So, [tex]\overline{CD}[/tex] bisects [tex]\overline{AB}[/tex].
Hence proved.