How is bisector of one side of a right angled triangle, drawn from right angled corner equal to the half of the bisected side?

Question

In a right angled triangle ABC with right angle at B and D being the mid-point of side AC, is it possible to prove BD=AD=CD without using co-ordinate geometry or circle theorems etc? (Just by using other elementary theorems: This has bothered me since my school years)

Thales' theorem: any right triangle can be inscribed in a semi-circle with the diameter coincident with the hypotenuse. — Bill Kleinhans, Oct 12 '13 at 20:17

André Nicolas · Answer 1 · 2013-10-12T04:18:22.140

2

A start: Complete the triangle to a rectangle in the obvious way. (Make a copy of the triangle, and rotate it into place.)

edited Oct 12 '13 at 04:18

answered Oct 12 '13 at 04:11

André Nicolas

507,029

Does it require to BDB' fall in same line? – user1077777777777777 Oct 12 '13 at 04:29
GOT it by ASA.. Cool – user1077777777777777 Oct 12 '13 at 04:32
Good. After one draws the rectangle, a bit of angle-chasing finishes things. – André Nicolas Oct 12 '13 at 04:42

How is bisector of one side of a right angled triangle, drawn from right angled corner equal to the half of the bisected side?

1 Answers1