Right triangle ABC is represented with right angle at vertex B. Points D and E lies on sides AB and BC. Segment DE is drawn.
In ΔABC,
and m∠ABC = 90°. D and E are the midpoints of
and
, respectively.
If the length of
is 9 units, the length of
is
units and m∠CAB is
Let's denote the length of AB as x.
Since D is the midpoint of AB, AD = BD = x/2.
Since E is the midpoint of BC, EC = CE = x/2.
Since the triangle is right-angled at B, by the Pythagorean theorem, we have AC^2 = AB^2 + BC^2.
Since AB = x and BC = x, we can substitute the values and get AC^2 = x^2 + x^2 = 2x^2.
Therefore, AC = √(2x^2) = x√2.
Since D and E are the midpoints of AB and BC respectively, DE is the median to AC.
By the midsegment theorem in triangles, the length of DE is half the length of AC, so DE = AC/2 = (x√2)/2 = x√2/2.
If the length of DE is 9 units, then we can set up the equation x√2/2 = 9 and solve for x.
x√2/2 = 9
x√2 = 18
x = 18/√2
x = 9√2
Therefore, the length of AB is 9√2 units.
Since AB = x = 9√2, and BD = x/2 = (9√2)/2 = 4.5√2, the length of BD is 4.5√2 units.
Since AB = x = 9√2, and EC = x/2 = (9√2)/2 = 4.5√2, the length of EC is 4.5√2 units.
Since AD = BD = 4.5√2 units, and EC = CE = 4.5√2 units, we can conclude that DE is a chord of a circle whose diameter is AC.