Mathematics - Properties of Triangle Question with Solution | TestHub

Explanation:

Let $△ABC$ be a triangle with side lengths $BC=a$, $CA=b$, and $AB=c$.

$AD$ is the internal bisector of $\angle A$, and $A{D}^{\prime }$ is the external bisector of $\angle A$.

The points $B,D,C,D^{\prime }$ are collinear.

1. Internal Angle Bisector Theorem:

For the internal bisector $AD$, it divides the side $BC$ in the ratio of the other two sides.

$$\frac{BD}{DC}=\frac{AB}{AC}=\frac{c}{b}$$

We know that $BD+DC=BC=a$.

From $\frac{BD}{DC}=\frac{c}{b}$, we have $BD=\frac{c}{b}DC$.

Substituting this into $BD+DC=a$:

$$\frac{c}{b}DC+DC=a$$

$$DC\left(\frac{c}{b}+1\right)=a$$

$$DC\left(\frac{c+b}{b}\right)=a$$

$$DC=\frac{ab}{c+b}$$

Then,

$$BD=a-DC=a-\frac{ab}{c+b}=\frac{a(c+b)-ab}{c+b}=\frac{ac+ab-ab}{c+b}=\frac{ac}{c+b}$$

So, option A is correct: $BD=\frac{ac}{c+b}$.

2. External Angle Bisector Theorem:

For the external bisector $A{D}^{\prime }$, it divides the side $BC$ externally in the ratio of the other two sides.

$$\frac{B{D}^{\prime }}{C{D}^{\prime }}=\frac{AB}{AC}=\frac{c}{b}$$

Since $D^{\prime }$ is on the line $BC$ externally, and $b<c$, $D^{\prime }$ will be on the side of $C$ away from $B$.

So, $B{D}^{\prime }=BC+C{D}^{\prime }=a+C{D}^{\prime }$.

Substituting this into $\frac{B{D}^{\prime }}{C{D}^{\prime }}=\frac{c}{b}$:

$$\frac{a+C{D}^{\prime }}{C{D}^{\prime }}=\frac{c}{b}$$

$$b(a+C{D}^{\prime })=c(C{D}^{\prime })$$

$$ab+b(C{D}^{\prime })=c(C{D}^{\prime })$$

$$ab=(c-b)C{D}^{\prime }$$

$$C{D}^{\prime }=\frac{ab}{c-b}$$

Then,

$$B{D}^{\prime }=a+C{D}^{\prime }=a+\frac{ab}{c-b}=\frac{a(c-b)+ab}{c-b}=\frac{ac-ab+ab}{c-b}=\frac{ac}{c-b}$$

So, option B is correct: $B{D}^{\prime }=\frac{ac}{c-b}$.

3. Distance between $D$ and $D^{\prime }$:

The points $B,D,C,D^{\prime }$ are collinear in that order (since $D$ is between $B$ and $C$, and $D^{\prime }$ is external to $BC$ on the side of $C$ given $b<c$).

Therefore, $D{D}^{\prime }=DC+C{D}^{\prime }$.

We found $DC=\frac{ab}{c+b}$ and $C{D}^{\prime }=\frac{ab}{c-b}$.

$$D{D}^{\prime }=\frac{ab}{c+b}+\frac{ab}{c-b}$$

$$D{D}^{\prime }=ab\left(\frac{1}{c+b}+\frac{1}{c-b}\right)$$

$$D{D}^{\prime }=ab\left(\frac{(c-b)+(c+b)}{(c+b)(c-b)}\right)$$

$$D{D}^{\prime }=ab\left(\frac{2c}{c^2-b^2}\right)$$

$$D{D}^{\prime }=\frac{2abc}{c^2-b^2}$$

So, option C is correct: $D{D}^{\prime }=\frac{2abc}{c^2-b^2}$.

4. Relationship between $AD$ and $BD$:

Option D states $AD>BD$. This is not generally true for all triangles. The length of the angle bisector $AD$ is given by the formula:

$$A{D}^2=bc\left(1-{\left(\frac{a}{b+c}\right)}^2\right)=bc-BD\cdot DC$$

We have $BD=\frac{ac}{c+b}$.

The inequality $AD>BD$ depends on the specific values of $a,b,c$. For example, in an equilateral triangle, $AD=\frac{\sqrt{3}}{2}a$ and $BD=\frac{a}{2}$, so $AD>BD$. However, this is not a universal property for all triangles. Therefore, option D is not always true.

The correct options are A, B, and C.

The final answer is $\boxed{A,B,C}$.