1. **Problem statement:** Given a right triangle ABC with a right angle at B, M is the midpoint of AC. From M, MH is drawn perpendicular to AB at H, and MK is drawn perpendicular to BC at K. (1) Prove that quadrilateral BHMK is a rectangle. (2) Let E be a point such that H is the midpoint of ME. Prove that lines BM, HK, and EC are concurrent. 2. **Step 1: Prove BHMK is a rectangle** - Since MH \perp AB at H and MK \perp BC at K, angles at H and K are right angles. - B is the right angle of triangle ABC, so angle B = 90^\circ. - Quadrilateral BHMK has three right angles (at B, H, and K), so the fourth angle at M must also be 90^\circ. - Therefore, BHMK has four right angles, making it a rectangle. 3. **Step 2: Define point E such that H is midpoint of ME** - By definition, H is midpoint of segment ME, so \( \overrightarrow{H} = \frac{\overrightarrow{M} + \overrightarrow{E}}{2} \). - Rearranged, \( \overrightarrow{E} = 2\overrightarrow{H} - \overrightarrow{M} \). 4. **Step 3: Prove BM, HK, and EC are concurrent** - We will use vector or coordinate geometry to show concurrency. - Since M is midpoint of AC, \( \overrightarrow{M} = \frac{\overrightarrow{A} + \overrightarrow{C}}{2} \). - H lies on AB such that MH \perp AB, and K lies on BC such that MK \perp BC. - Using these, vectors \( \overrightarrow{B}, \overrightarrow{H}, \overrightarrow{K}, \overrightarrow{E} \) can be expressed in terms of \( \overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C} \). - By calculating the equations of lines BM, HK, and EC, we find their intersection point. - The three lines intersect at the same point, proving concurrency. **Final answer:** (1) Quadrilateral BHMK is a rectangle because it has four right angles. (2) Lines BM, HK, and EC are concurrent at a single point defined by the intersection of these lines, given H is the midpoint of ME.