Question

in triangle abc angle b = 90 degree P is the midpoint of the hypotenuse AC. prove that BP = half of AC.

Answer 1

By theorem and perpendicular drawn from a right angle of a right angle triangle when the line is perpendicular to the hypotenuse we get that DP is a perpendicular bisector of faces since it is a perpendicular bisector ac. equals to PC

Answer 2

Answer :

Let, triangle ABC be a right angle triangle at ∠B

Let P be the mid-point of hypotenuse AC

Draw a circle with centre P and AC as diameter

Since,

∠ABC = 90°

Therefore, the circle passes through B

Therefore,

BP = Radius

Also,

AP = CP = Radius

Therefore,

AP = BP = CP

Hence, BP = ½ AC