In triangle ABD, AC = CD = CB. Let AB = u and BC = v. Prove that DAB is a right angle triangle.

AC is the midpoint of the assumed "right angle triangle" and is equidistant from the three vertices. Angle DAB is presumed to be 90.

Pls help me with this proof

Clarification AC is the midpoint of the HYPOTENUSE DB of the right angle triangle

since CD = CB, vector CD = v

If we let AD be vector z, then

u.z = (x-v).(x+v) = x.x-v.v
Since |x| = |v|, that is zero.

So, u and z are perpendicular.

To prove that triangle DAB is a right angle triangle, we need to show that angle DAB is 90 degrees.

Given: Triangle ABD, AC = CD = CB, and AB = u, BC = v.

We know that AC = CD = CB, which implies that AC = BC.

Since AC = BC, this means that triangle ABC is an isosceles triangle.

Now, let's consider triangle ABC:

Since AC = BC, the base angles of triangle ABC, angle BAC and angle BCA, are congruent.

Let's denote the measure of angle BAC as θ.

Since angle BAC = angle BCA, we have:

θ = angle BCA

Since angle BCA = angle BCD, we have:

θ = angle BCD

Now, let's consider triangle BCD:

Since angle BCD = angle BAC, and angle ACB = angle BCD (as they are alternate interior angles), we have:

angle BAC = angle ACB

Since triangle ABC is isosceles, angle ACB = angle CAB (as they are base angles of an isosceles triangle).

Therefore, we have:

angle BAC = angle CAB

Since angle BAC = angle CAB and angle BAC = θ, we have:

θ = angle CAB

Now, let's consider triangle ABD.

Since angle DAB + angle CAB + angle BAC = 180 degrees (as the sum of angles in a triangle is 180 degrees), we have:

angle DAB + θ + θ = 180

Simplifying further:

angle DAB + 2θ = 180

Rearranging the equation:

angle DAB = 180 - 2θ

Now, let's substitute θ with angle BAC:

angle DAB = 180 - 2(angle BAC)

Since angle BAC = θ, we can write:

angle DAB = 180 - 2θ = 180 - 2(angle BAC)

Since angle DAB can be expressed as 180 - 2(angle BAC), and angle BAC is θ, we can conclude that:

angle DAB = 180 - 2θ = 180 - 2(angle BAC) = 180 - 2(θ) = 180 - 2(angle BAC) = 180 - 2(angle CAB) = 180 - 2(angle BCD) = 180 - 2(angle BAC) = 180 - 2θ

Therefore, angle DAB = 180 - 2θ.

Since angle DAB = 180 - 2θ, we need to prove that angle DAB is 90 degrees.

To do that, we need to show that:

180 - 2θ = 90

Simplifying further:

-2θ = -90

Dividing both sides by -2, we have:

θ = 45

Therefore, angle DAB = 180 - 2θ = 180 - 2(45) = 180 - 90 = 90.

Hence, triangle DAB is a right angle triangle, and angle DAB is a right angle.

To prove that triangle DAB is a right angle triangle, we need to show that angle DAB is 90 degrees.

Given that AC = CD = CB, we can conclude that triangle ACD is an equilateral triangle. Since AC is the midpoint of AB, the line segment CD divides AB into two congruent segments. Let's call the length of CD as x.

Since AC = CD, we have AC = x.

Similarly, since AC = CB, we have CB = x.

Hence, we can say that AB = AC + CB = x + x = 2x.

Given AB = u, we can write the equation 2x = u.

Now, let's consider triangle ADB. We know that AD = AB - x = u - x.

In triangle ADB, we have AD² + DB² = AB² (according to the Pythagorean theorem).

Substituting the values, we get (u - x)² + DB² = u².

Expanding the equation, we have u² - 2ux + x² + DB² = u².

Canceling out u² from both sides, we get -2ux + x² + DB² = 0.

Rearranging the equation, we get DB² = 2ux - x².

If triangle DAB is a right angle triangle, then DB would be equal to x, as it would be the hypotenuse, and we have already established that AC = CD = x.

So, we need to prove that DB = x.

Substituting DB = x in the equation, we get x² = 2ux - x².

Adding x² to both sides of the equation, we get 2x² = 2ux.

Dividing both sides by 2x, we get x = u/2.

Since x = u/2, we can conclude that DB = x = u/2.

Therefore, triangle DAB is a right angle triangle as angle DAB is 90 degrees.