In the isosceles triangle ABC, we assume that it is sides AC and BC that

are equal.

angle CDA = angle CDB and both are right angles, as CD is an altitude

from AB to C.

angle CAD = angle CBD as they are the angles opposite the equal sides

of an isosceles triangle.

angle ACD = angle DCB as these are the third angles in two triangles whose

other two angle are equal and so equal because the angle sum of any

triangle is two right angles.

Therefore as side CD is common to both triangle ACD and triangle BCD

these triangles are congruent by ASA.

Hence AD is congruent to DC as these are corresponding sides of congruent

triangles. So we have proven that D bisects AC.

RonL