Prove $BD=2CD$ in Triangle $ABC$

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In summary, to prove that $BD=2CD$ in Triangle $ABC$, you can use the properties of similar triangles or the triangle inequality theorem. It is also possible to use the Pythagorean theorem to prove this relationship. This concept is closely related to the concept of ratios, where we compare the lengths of $BD$ and $CD$ in a ratio of $2:1$. However, this relationship is not a necessary and sufficient condition for Triangle $ABC$ to be isosceles.
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In a triangle $ABC$, it's given that $AB=AC$, point $D$ is on $BC$ whereas point $E$ is on $AD$ such that $\angle BED=2 \angle CED=\angle BAC$.

Prove that $BD=2CD$.

Note:

This problem is actually posted by Albert at another math forum about 2 years ago and for all information, I have gained his "permission" to post the exact same problem here, for the folks to have some fun solving it either trigonometrically or geometrically.
 

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My solution:
View attachment 2541

Since AB=AC, I first let $\angle CED=\alpha$, then $\angle EBC=\beta$ and $\angle DAC=\theta$.
A little working on the triangle $EBC$ shows that $\alpha+\beta+\theta=90^{\circ}$

Consider the triangles $EBD$ and $ECD$, we have:
$\dfrac{ED}{\sin \beta}=\dfrac{BD}{\sin 2\alpha}$ and $\dfrac{ED}{\sin(\beta-\alpha+2\theta)}=\dfrac{CD}{\sin \alpha}$

This yields
$\dfrac{BD}{CD}=\left(\dfrac{\sin 2\alpha}{\sin \alpha} \right)\left(\dfrac{\sin (\beta-\alpha+2\theta)}{\sin \beta} \right)$

We then try to eliminate the variable $\beta$ by using the relation $\beta=90^{\circ}-(\alpha+\theta)$, we get:

$\begin{align*}\dfrac{BD}{CD}&=\left(\dfrac{\sin 2\alpha}{\sin \alpha} \right) \left(\dfrac{\sin (90^{\circ}-(\theta+\alpha))-\alpha+2\theta)}{\sin (90^{\circ}-(\theta+\alpha))} \right)\\&=\dfrac{\sin 2\alpha}{\sin \alpha}\cdot\dfrac{\sin (90^{\circ}-(2\alpha-\theta))}{\sin(90^{\circ}-(\theta+\alpha))}\\&=\dfrac{\sin 2\alpha}{\sin \alpha}\cdot\dfrac{\cos (2\alpha-\theta)}{\cos (\theta+\alpha)}\\&=\dfrac{\sin 2\alpha}{\sin \alpha}\cdot\dfrac{\cos 2\alpha \cos\theta+\sin 2\alpha \sin \theta}{\cos \theta \cos\alpha-\sin \theta \sin \alpha}(*)\end{align*}$

We know the RHS expression will be reduced to 2, so, I think eliminating another variable would be a wise way to go.

Now, consider the triangles $ABE$ and $ACE$, we have:

$\dfrac{AB}{\sin(180^{\circ}- 2\alpha)}=\dfrac{AE}{\sin \theta}$ and $\dfrac{AC}{\sin(180^{\circ}- \alpha)}=\dfrac{AE}{\sin(\alpha-\theta)}$

Since $AB=AC$, the above simplifies to:

$\dfrac{\sin 2\alpha}{\sin \theta}=\dfrac{\sin \alpha}{\sin(\alpha-\theta)}$

$\dfrac{2\sin \alpha \cos \alpha}{\sin \alpha}\cdot\sin(\alpha-\theta)=\sin \theta$

$2\cos \alpha\cdot\sin(\alpha-\theta)=\sin \theta$

$2\cos \alpha\cdot(\sin \alpha \cos\theta-\cos \alpha \sin\theta)=\sin \theta$

$2\cos \alpha\cdot\sin \alpha \cos\theta=\sin \theta(1+2\cos ^2 \alpha)$

$\tan \theta=\dfrac{\sin 2\alpha}{1+2\cos ^2 \alpha}$ (**)

To merge these two equations ((**)and (*)), we divide the equation (*), top and bottom, by $\cos \theta$ to get:

$\dfrac{BD}{CD}=\dfrac{\sin 2\alpha}{\sin \alpha}\cdot \dfrac{\cos 2\alpha +\sin 2\alpha \tan\theta}{\cos \alpha -\sin \alpha \tan\theta}$

therefore,

$\begin{align*}\dfrac{BD}{CD}&=\dfrac{2\sin \alpha \cos \alpha}{\sin \alpha}\cdot \dfrac{\cos 2\alpha +\sin 2\alpha\cdot\frac{\sin 2\alpha}{1+2\cos ^2 \alpha}}{\cos \alpha-\sin \alpha\left(\dfrac{\sin 2\alpha}{1+2\cos ^2 \alpha} \right)}\\&=2 \cos \alpha\cdot \dfrac{\cos 2\alpha +2\cos^2 \alpha \cos 2\alpha+\sin^2 2\alpha}{\cos \alpha+2\cos^3 \alpha-\sin \alpha \sin 2\alpha}\\&=2 \cos \alpha\cdot \dfrac{\cos 2\alpha(1+2\cos^2 \alpha)+\sin^2 2\alpha}{\cos \alpha+2\cos^3 \alpha-2\sin \alpha \sin \alpha \cos \alpha}\\&=2 \cos \alpha\cdot \dfrac{\cos 2\alpha(\cos 2\alpha+2)+\sin^2 2\alpha}{\cos \alpha(1+2\cos^2\alpha-2\sin^2 \alpha)}\\&=2 \cos \alpha\cdot \dfrac{\cos^2 2\alpha+\sin^2 2\alpha+2\cos 2\alpha}{\cos \alpha(1+2(\cos^2\alpha-\sin^2 \alpha))}\\&=2 \cos \alpha\cdot \dfrac{1+2\cos 2\alpha}{\cos \alpha(1+2\cos 2\alpha)}\\&=2 \cos \alpha\cdot\dfrac{1}{\cos \alpha}\\&=2 \end{align*}$

Hence,

$BD=2CD$ (Q.E.D.)
 

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FAQ: Prove $BD=2CD$ in Triangle $ABC$

How do you prove that $BD=2CD$ in Triangle $ABC$?

To prove that $BD=2CD$ in Triangle $ABC$, you can use the properties of similar triangles. Since $BD$ and $CD$ share a common vertex $C$, we can draw a line from $B$ to $D$ that is parallel to $AC$. This creates two similar triangles, $ABC$ and $BCD$. Using the property that corresponding sides of similar triangles are proportional, we can set up the following equation: $\frac{BD}{CD}=\frac{BC}{AC}=\frac{2}{1}$. This shows that $BD$ is indeed twice the length of $CD$.

Can you prove that $BD=2CD$ without using similar triangles?

Yes, it is possible to prove $BD=2CD$ without using similar triangles. One way to do this is by using the triangle inequality theorem. This theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Since we know that $AC>CD$, we can use the triangle inequality theorem to show that $AC+CD>CD+CD$, which simplifies to $AC>2CD$. Similarly, we can show that $AB>2CD$. Combining these two inequalities, we get $AB+AC>2CD+2CD$, which simplifies to $AB+AC>4CD$. Finally, using the triangle inequality theorem again, we can show that $AB+AC>BD$, which proves that $BD$ is indeed twice the length of $CD$.

Is $BD=2CD$ a necessary and sufficient condition for Triangle $ABC$ to be isosceles?

No, $BD=2CD$ is not a necessary and sufficient condition for Triangle $ABC$ to be isosceles. To show this, we can consider an equilateral triangle $ABC$. In this triangle, all three sides are equal, including $BD$ and $CD$, but the triangle is not isosceles. Therefore, $BD=2CD$ is not enough to prove that Triangle $ABC$ is isosceles. However, if we also know that $AB=AC$, then $BD=2CD$ would be a sufficient condition for Triangle $ABC$ to be isosceles.

Can you use the Pythagorean theorem to prove $BD=2CD$ in Triangle $ABC$?

Yes, it is possible to use the Pythagorean theorem to prove $BD=2CD$ in Triangle $ABC$. We can draw a perpendicular line from $B$ to $AC$, creating two right triangles, $ABC$ and $BCD$. Using the Pythagorean theorem, we can set up the following equation for each triangle: $AB^2=BC^2+AC^2$ and $BD^2=CD^2+BC^2$. Since we know that $AB=2AC$ and $AB=BD+CD$, we can substitute these values into the equations and solve for $BD$ and $CD$. This will show that $BD=2CD$, proving the desired result.

How does proving $BD=2CD$ in Triangle $ABC$ relate to the concept of ratios?

Proving $BD=2CD$ in Triangle $ABC$ is closely related to the concept of ratios. In this case, we are comparing the lengths of $BD$ and $CD$ to show that they are in a ratio of $2:1$. This is similar to comparing the sides of similar triangles, where we also use ratios to show the proportional relationship between corresponding sides. Additionally, the fact that $BD=2CD$ in Triangle $ABC$ can also be interpreted as the ratio of the whole (the hypotenuse $AB$) to one of its parts (the leg $AC$) being equal to the ratio of the same part (the leg $BC$) to the other part (the leg $CD$). This is a fundamental concept in the study of ratios and their applications.

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