Prove that the straight line joining the middle point of the hypotenuse of a right angled triangle to the right angle is equal to half the hypotenuse. I am supposed to use the following 8 theorems only to prove the above prepositions:

Theorem 1: If a ray stands on a line , then the sum of the adjacent angles formed is $180 $deg.

Theorem 2: If two lines intersect , then the vertically opposite angles are equal.

Theorem 3: If a transversal cuts two parallel lines, then each pair of alternate angles are equal, and the interior angles on the same side of the transversal are supplementary.

Theorem 4: Lines which are parallel to the same line are parallel to each other.

Theorem 5: The sum of the three angles of a triangle is $180$deg.

Theorem 6: If one side of a triangle is produced , the exterior angle so formed is equal to the sum of the interior opposite angles.

Theorem 7: The angles opposite to equal sides of a triangle are equal in an isosceles triangle.

Theorem 8: The bisector of the vertical angle of an isosceles triangle bisects the base and is perpendicular to the base.

I tried to get the solution but could not apply theorems 7 and 8 since it is a right angle triangle. I don't use concept of rotation. I use ASA, SAS, SSS and RHS Postulates and converse of theorem 1,3,7 and 8. can anyone tell me how the proof looks like? Thanks in advance...Srikanth
in $ \triangle ABC$, $\angle B=90^\circ$, $AD=DC$


we draw $DO \perp BC$. So $BO=OC= \frac{BC}{2}$ and $DO= \frac{AB}{2} $ [midpoint theorem]
so $ BD^2=DO^2+BO^2=\frac{BC^2+AB^2}{4}=\frac{AC^2}{4}$
so $BD= \frac{AC}{2}$ (proved)
I didn't read Theorems 1–8, but offer the following proof: The midpoint $M$ of the hypotenuse $AB$ is the center of a Thales circle with  diameter $AB$, and $MC$ is a radius of this circle.