Suppose your right triangle is ABC, with C being the vertex with the right angle. Suppose we call the midpoint of hypotenuse AB point M. Then line segment CM is the one you are trying to show is half the length of AB.
A segment drawn from M perpendicular to BC will intersect the midpoint of BC. Call that point P. The right triangles thus formed (CMP and BMP) are congruent, so the median line CM must be the same length as half the hypotenuse.
Several theorems must be invoked to show the triangles are congruent. Segment MP will be parallel to AC, so the angle with the hypotenuse (PMB) must match the angle at the end of the parallel leg (CAB). This means the triangles ABC and MBP are similar. The segments in similar triangles have the same scale factor, so the fact that MB is half of AB means that PB is half of CB. That means length CP is equal to length PB, and triangles MCP and MBP must be congruent. They have two angles the same (MPC = MPB = 90 degrees), and the segments on either side of those angles are the same (MP=MP, and PC = PB). The Pythagorean Theorem requires MC = MB, completing your proof.
A segment drawn from M perpendicular to BC will intersect the midpoint of BC. Call that point P. The right triangles thus formed (CMP and BMP) are congruent, so the median line CM must be the same length as half the hypotenuse.
Several theorems must be invoked to show the triangles are congruent. Segment MP will be parallel to AC, so the angle with the hypotenuse (PMB) must match the angle at the end of the parallel leg (CAB). This means the triangles ABC and MBP are similar. The segments in similar triangles have the same scale factor, so the fact that MB is half of AB means that PB is half of CB. That means length CP is equal to length PB, and triangles MCP and MBP must be congruent. They have two angles the same (MPC = MPB = 90 degrees), and the segments on either side of those angles are the same (MP=MP, and PC = PB). The Pythagorean Theorem requires MC = MB, completing your proof.
