Mathematics - Properties of Triangle Question with Solution | TestHub

Problem: If $l$ is the median from the vertex $A$ to the side $BC$ of a $△ABC$, then

Options:

A. $4l^2=2b^2+2c^2-a^2$

B. $4l^2=b^2+c^2+2bc\cos A$

C. $4l^2=a^2+4bc\cos A$

D. $4l^2=(2\mathrm{s}-\mathrm{a}{)}^2-4\mathrm{bc}{\sin }^2\left(\frac{A}{2}\right)$

Correct Answer: A

Explanation:

Let $△ABC$ be a triangle with side lengths $a,b,c$ opposite to vertices $A,B,C$ respectively. Let $l_a$ (denoted as $l$ in the question) be the length of the median from vertex $A$ to side $BC$. This median bisects side $BC$ at point $D$, so $BD=DC=\frac{a}{2}$.

We can use Apollonius's Theorem to relate the lengths of the sides of a triangle to the length of a median.

Apollonius's Theorem states that for a triangle $ABC$ with median $AD$ to side $BC$:

$A{B}^2+A{C}^2=2(A{D}^2+B{D}^2)$

In our notation:

$c^2+b^2=2(l^2+(\frac{a}{2}{)}^2)$

$c^2+b^2=2(l^2+\frac{a^2}{4})$

$c^2+b^2=2l^2+\frac{a^2}{2}$

To solve for $l^2$, we can rearrange the equation:

$2l^2=c^2+b^2-\frac{a^2}{2}$

Multiply by 2 to clear the fraction:

$4l^2=2c^2+2b^2-a^2$

This matches option A.

Let's analyze the other options using the Law of Cosines.

The Law of Cosines states:

$a^2=b^2+c^2-2bc\cos A$

$b^2=a^2+c^2-2ac\cos B$

$c^2=a^2+b^2-2ab\cos C$

Consider option B: $4l^2=b^2+c^2+2bc\cos A$

From Apollonius's Theorem, we have $4l^2=2b^2+2c^2-a^2$.

Substitute $a^2=b^2+c^2-2bc\cos A$ into the expression for $4l^2$:

$4l^2=2b^2+2c^2-(b^2+c^2-2bc\cos A)$

$4l^2=2b^2+2c^2-b^2-c^2+2bc\cos A$

$4l^2=b^2+c^2+2bc\cos A$

This shows that option B is also correct.

Consider option C: $4l^2=a^2+4bc\cos A$

From option B, we have $4l^2=b^2+c^2+2bc\cos A$.

We know $a^2=b^2+c^2-2bc\cos A$.

So, $b^2+c^2=a^2+2bc\cos A$.

Substitute this into the expression for $4l^2$:

$4l^2=(a^2+2bc\cos A)+2bc\cos A$

$4l^2=a^2+4bc\cos A$

This shows that option C is also correct.

Consider option D: $4l^2=(2\mathrm{s}-\mathrm{a}{)}^2-4\mathrm{bc}{\sin }^2\left(\frac{A}{2}\right)$

Here, $s$ is the semi-perimeter, $s=\frac{a+b+c}{2}$.

So, $2s=a+b+c$.

Then $2s-a=b+c$.

The expression becomes $4l^2=(b+c{)}^2-4bc{\sin }^2(\frac{A}{2})$.

We know that ${\sin }^2(\frac{A}{2})=\frac{1-\cos A}{2}$.

So, $4bc{\sin }^2(\frac{A}{2})=4bc\frac{1-\cos A}{2}=2bc(1-\cos A)=2bc-2bc\cos A$.

Substitute this back into the expression:

$4l^2=(b+c{)}^2-(2bc-2bc\cos A)$

$4l^2=b^2+2bc+c^2-2bc+2bc\cos A$

$4l^2=b^2+c^2+2bc\cos A$.

This is the same as option B, which we already established as correct. Therefore, option D is also correct.

Since the question asks "then", implying one correct option, and all options A, B, C, D are mathematically equivalent expressions for $4l^2$, there might be an issue with the question expecting a single answer or it's a multiple-correct-answer type question. However, typically in such problems, Apollonius's Theorem (Option A) is the most direct and fundamental relationship.

Final Answer is A, B, C, D. All options are equivalent expressions for $4l^2$.

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