In Fig. 14.40, a right triangle `B O A` is given. `C` is the mid-point of the hypotenuse `A B` . Show that it is equidistant from the vertices `O ,\ A` and `B` . (FIGURE)

Prove that the mid-point of the hypotenuse of a right triangle is equidistant from its vertices.

Prove that the mid-point of the hypotenuse of right angled triangle is equidistant from its vertices.

Prove that the mid-point of the hypotenuse of right angled triangle is equidistant from its vertices.

Prove that the mid-point of the hypotenuse of right angled triangle is equidistant from its vertices.

If D is the mid-point of the hypotenuse A C of a right triangle A B C , prove that B D=1/2A C

ABC is a triangle. Locate a point in the interior of DeltaA B C which is equidistant from all the vertices of DeltaA B C

The coordinates of the point which is equidistant from the points O(0,0,0) A (a,0,0), B(0,b,0) and C(0,0,c)

In Figure, line segments A B is parallel to another line segment C DdotO is the mid-point of A Ddot Show that: O is also the mid-point of B D

In right triangle A B C , right angle at C ,\ M is the mid-point of the hypotenuse A Bdot C is jointed to M and produced to a point D such that D M=C Mdot Point D is joined to point Bdot Show that A M C~= B M D (ii) /_D B C=/_A C B D B C~= A C B (iii) C M=1/2A B

Prove by vector method, that in a right-angled triangle ABC, AB^(2) + AC^(2) = BC^(2) , the angle A being right angled. Also prove that mid-point of the hypotenuse is equidistant from vertex.