Prove $(CD)^2=(BD)^2+\dfrac{(BC)^2(AD)}{AB}$ for Isosceles Trigon ABC

[solakis1]

Given an isosceles trigon ABC (AC=AB),take apoint on AB ,D and between the points A and B,then prove:

$(CD)^2=(BD)^2+\dfrac{(BC)^2(AD)}{AB}$

 

[jvicens]

Hi there,

Thank you for sharing this interesting problem! I am always excited to delve into mathematical proofs and explore the beauty of geometry.

To begin, let us label the given triangle as ABC, with point D on the side AB (as shown in the diagram below).

Now, we can use the Pythagorean Theorem to find the lengths of segments BC and BD.

According to the theorem, in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In triangle BCD, we have:

$(CD)^2=(BC)^2+(BD)^2$

Next, let's take a look at triangle ACD. Since AC=AB, we can conclude that triangle ACD is also an isosceles triangle. This means that the base angles, ∠ACD and ∠ADC, are equal.

Now, let's use the property of isosceles triangles that states the base angles are equal to the opposite sides. This means that AC=CD.

Substituting this into our equation, we get:

$(CD)^2=(BC)^2+(BD)^2=(BC)^2+(AC)^2$

From the given information, we know that $AC=AB$, so we can substitute $AB$ for $AC$ in the equation above.

$(CD)^2=(BC)^2+(AB)^2$

Now, let's look at triangle ABD. By using the Pythagorean Theorem again, we can find the length of segment AD.

$(AD)^2=(AB)^2+(BD)^2$

Solving for $(AD)$, we get:

$(AD)=(AB)^2-(BD)^2$

Substituting this into our original equation, we get:

$(CD)^2=(BC)^2+(AB)^2=(BC)^2+[(AB)^2-(BD)^2]$

Simplifying, we get:

$(CD)^2=(BC)^2+(AB)^2-(BD)^2$

Now, let's go back to the original equation we were trying to prove:

$(CD)^2=(BD)^2+\dfrac{(BC)^2(AD)}{AB}$

Substituting $(AD)$ from our previous equation, we get:

$(CD)^2=(BD)^2+\dfrac{(BC)^2[(AB)^2-(BD