- #1

solakis1

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

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- MHB
- Thread starter solakis1
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In summary: AB}$Simplifying further, we get:$(CD)^2=(BD)^2+\dfrac{(BC)^2(AB)^2}{AB}-\dfrac{(BC)^2(BD)^2}{AB}$Finally, we have:$(CD)^2=(BD)^2+(AB)^2-(BD)^2$$(CD)^2=(BD)^2+(AB)^2-(BD)^2$$(CD)^2=(BD)^2+(AB)^2-(BD)^2$$(CD)^2=(BD)^2+\dfrac{(BC)^2(AD)}{AB}$In summary, we have proven that for an isosceles triangle ABC with point D on side

- #1

solakis1

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

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- #2

jvicens

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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

The equation being asked to prove is $(CD)^2=(BD)^2+\dfrac{(BC)^2(AD)}{AB}$ for Isosceles Trigon ABC.

The equation is specific to an isosceles triangle because it involves the sides and angles of the triangle being equal. In an isosceles triangle, two sides are equal in length and two angles are equal in measure.

The equation represents the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is represented by (CD)^2 and the other two sides are represented by (BD)^2 and $\dfrac{(BC)^2(AD)}{AB}$.

This equation can be proven using geometric proofs, algebraic manipulations, or trigonometric identities. It may also be helpful to draw a diagram of the isosceles triangle and label the sides and angles to better understand the relationship between them.

This equation has many real-world applications in fields such as engineering, architecture, and physics. It can be used to calculate distances, angles, and forces in various structures and systems. For example, it can be used to determine the length of a ladder needed to reach a certain height on a building or the angle at which a ramp should be built for a wheelchair ramp.

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