Answer:
In \(\triangle shd\) and \(\triangle std\)
\(sd \perp ht\)
By definition of perpendicular: Lines that are right angle to each other.
\(\angle sdh \cong \angle sdt = 90^{\circ}\) [Right angle]
\(sh \cong st\) [Hypotenuse side] [Given]
Transitive property says that segment is congruent to itself.
\(sd \cong sd\) [By transitive property]
HL theorem states that if the hypotenuse and one leg of right triangle are equal to the hypotenuse and one leg of the another right angle , then the two triangle are congruent.
then by HL theorem;
\(\triangle shd \cong \triangle std\) proved!