given: sd⊥ht; sh≅st
prove: δshd = δstd

The complete problem is attached.

According to the problem, SD is perpendicular to HT and SH is congruent to ST. Also, SD is a common side for both right triangles. Let's demonstrate with the same order shown'

                Statement                                  Reason

1. SD ⊥ HT                                               1. Given

2. ∠SDH and ∠SDT are right ∠s.           2. By definition of perpendicularity.

3. SH ≅ ST                                              3. Given.

4. SD ≅ SD.                                            4. By reflexive property.

5. ΔSHD ≅ ΔSTD.                                  5. By HL postulate.

The second statement is deducted by definition of perpendicularity, if SD ⊥ HT, that means those sides makes angles of 90°, that's why ∠SDH and ∠SDT are right ∠s.

The fourth statement is true by reflexive property, this property states that every side is equal to itself, so SD ≅ SD.

Lastly, the fifth statement is the final congruence between both triangles. Through those statements, it was demonstrate it that those triangles where right triangles, so the HL postulate is applied to right triangles only, because it states that if two right triangles have congruent hypothenuses and one congruent leg, then those right triangles are congruent.

okay wait before i answer, is this for your points on a graph?