Answer
The proof is in the following form:
Column 1 of proof; Column 2 of proof.
Thus, we have:
1. The altitude of a right triangle is orthogonal to the base; this is the definition of an altitude.
2. The right angles are congruent; right angles are always congruent to each other
3. The angle shared by both triangles is congruent; identity property.
4. Thus, the two right triangles are similar; AA
Work Step by Step
1- in a triangle ABC, $ \angle A + \angle B = 90^{\circ} $ since its a right triangle.
2-in a triangle ADC, $ \angle A + \angle ACD = 90^{\circ} $ same reason as step one.
3- by angle addition property, $ \angle A + \angle B= \angle A + \angle ACD $ therefore $ \angle B= \angle ACD $.
4- $ \angle B + \angle BCD = 90^{\circ} $ and $ \angle A + \angle ACD = 90 ^{\circ} $
5- by substitution, and since $ \angle ACD = \angle B $ in step three, therefore $ \angle A= \angle BCD$
6- triangle ACD ~ triangle CBD by AA.
7- $ \angle B = \angle B $ and $ \angle A = \angle A$ identity.
8- $ \angle ADC = \angle BDC = ACB $ right angles are congruent.
9- triangle ACD ~ triangle CBD ~ triangle ABC
$ \square $
