construct perpendicular bisectors on the two shorter sides (non hypotenuse)
the bisectors will intersect the hypotenuse at the vertices of the two isosceles triangles
the common side for the triangles is the line from the intersection to the right angle
How to divide a right angled triangle into two isosceles triangle ? Kindly help
2 answers
or even simpler:
draw a median from the right-angle to the hypotenuse.
proof:
let the triangle be A(0,0), B(b,0) and C(0,c)
let AD be the median.
midpoint of BC = ( (b/2 , c/2)
AD = √(b^2/4 + c^2/4)
BD = √(b - b/2)^2 + (c/2-0)^2 ) = CD
= √( b^2/4 + c^2/4)
= AD
draw a median from the right-angle to the hypotenuse.
proof:
let the triangle be A(0,0), B(b,0) and C(0,c)
let AD be the median.
midpoint of BC = ( (b/2 , c/2)
AD = √(b^2/4 + c^2/4)
BD = √(b - b/2)^2 + (c/2-0)^2 ) = CD
= √( b^2/4 + c^2/4)
= AD
1 answer