In RT triangle ABC, ad is the middle line on hypotenuse BC. How to prove ad = 1 / 2BC?

In RT triangle ABC, ad is the middle line on hypotenuse BC. How to prove ad = 1 / 2BC?

Draw a circle with BC side as diameter and midpoint D as center. Take any point a on the circle, connect AC and AB, then the angle BAC = 90 degrees, and then connect ad, then ad = CD = BD = circle radius, so ad = 1 / 2BC

CD is the high line on the hypotenuse of RT △ ABC, ad and BD are two of the equations x2-6x + 4 = 0, then the area of △ ABC is______ ．

∵ AD and BD are two of the equations x2-6x + 4 = 0, ∵ AD + BD = 6, ad · BD = 4, ∵ ACB = 90 °, CD ⊥ AB in D, ∵ DBC ∽ DCA, ∵ CDAD = DBCD, ∵ CD2 = ad · BD, ∵ CD = ad × BD = 2, ∵ s △ ABC = 12 × (AD + BD) × CD = 6

If BC = 10, ad bisects ∠ BAC, intersects BC at point D, and BC: CD = 3:2, then the distance from point d to line AB is
If BC = 10, ad bisects ∠ BAC and intersects BC at point D, and BD: CD = 3, then the distance from point d to line AB is
Wrong number in front, not BC: CD = 3: BC: CD = 3:2

Through point D, make de vertical AB, because ad bisector angle BAC, BD / CD is 3 / 2, so de = DC = 4

At RT △ ABC, ∠ C = 90 °, ad bisection ∠ BAC intersects BC at point D, and BD: CD = 3:2, BC = 15, calculate the distance from point d to line ab

Make the intersection of AB and e through point D
De is the required distance
It can be proved that triangle ADC is equal to triangle AED
So there is de = DC
DC = 15/5 * 2 = 6
So de = 6