Find Angle ∢CDB in an Isosceles Right Triangle with Median

ABC is an isosceles right triangle.

AB = AC

BD is the median of the triangle.

What is the size of the angle $∢CDB$ ?

Video Solution

Solution Steps

00:00 Determine the angle CDB

00:03 In an isosceles triangle, base angles are equal

00:10 The sum of angles in a triangle equals 180

00:13 Therefore subtract from 180 and divide by 2 in order to determine the angles

00:22 The values of angles A and C

00:31 BD is median according to the given data

00:34 The median to the hypotenuse in a right triangle equals half the hypotenuse

00:42 Therefore the median itself creates isosceles triangles

00:47 In isosceles triangles, base angles are equal

01:01 The sum of angles in a triangle equals 180

01:10 Therefore subtract the base angles from 180 to find CDB

01:15 That's the solution

Step-by-Step Solution

To solve this problem, we need to analyze the properties of the isosceles right triangle $\triangle ABC$ :

The triangle is isosceles and right ( $\angle BAC = 90^\circ$ ), meaning $\angle ABC = \angle ACB = 45^\circ$ .
$BD$ is the median, dividing the hypotenuse $AC$ into two equal parts ( $AD = DC$ ).

Now, we consider triangle $\triangle CDB$ :

Since $D$ is the midpoint of $AC$ , $AD = DC$ and $BD$ is the median, which also acts as the altitude because the triangle is isosceles and right.
Because $BD$ is an altitude in right triangle $\triangle ABC$ , triangle $\triangle CDB$ is itself an isosceles right triangle, with $\angle DBC = \angle DCB$ .

Using the properties of triangle $\triangle CDB$ :

The angle sum property of a triangle states that the sum of the angles in a triangle is $180^\circ$ .
$\angle CDB$ in triangle $\triangle CDB$ must therefore be $90^\circ$ , as $\angle DBC = \angle DCB$ are both $45^\circ$ .

Therefore, the size of $\angle CDB$ is $90^\circ$ .