Question

Let$D$ be the middle point of the side $BC$of a triangle $\vartriangle ABC$_.if the triangle $\vartriangle ADC$_{is equilateral, then ${{a}^{2}}:{{b}^{2}}:{{c}^{2}}$ is equal to}
A.$1:4:3$
B. $4:1:3$
C. $4:3:1$
D. $3:4:1$

Answer 1

Hint:
To solve this question, we will find all the angles of the triangle $\vartriangle ABC$ _{using properties and} theorems of the triangle. After this we will use Law of sine $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$and equate it some constant and by substituting all the values of the angles we will find the value of $a,b$and $c$. We will put the value of $a,b$and $c$ in ${{a}^{2}}:{{b}^{2}}:{{c}^{2}}$_{and then find} the ratio.
Complete step-by-step solution:
We are given a triangle $\vartriangle ABC$_{in which $D$} is the middle point on the side $BC$. We have to find the ratio ${{a}^{2}}:{{b}^{2}}:{{c}^{2}}$_{if the triangle $\vartriangle ADC$is equilateral.
In triangle $\vartriangle ABC$, the value of the sides will be,
$AB=c,BC=a$ and $AC=b$}.
_{Now we will make the diagram of the} triangle $\vartriangle ABC$ _{with $D$} being the middle point on the side $BC$forming an equilateral triangle $\vartriangle ADC$<sub>.



Because the triangle $\vartriangle ADC$is equilateral each of the angles in this triangle will be ${{60}^{0}}$ and all the sides will be also equal. That is
$AD=AC=DC$</sub>
And as $D$is the mid-point of side $BC$ so,
$BD=DC$
From here we can say that in triangle $\vartriangle ABD$_{, the sides \[BD\] and $AD$will} be equal, that is $BD=AD$ making it an isosceles triangle. And by the theorem of an isosceles triangle we know that angles opposite to the equal sides of an isosceles triangle are also equal.
Hence,
$\angle ABD=\angle DAB$
Let this angle be $x$ so,
$\angle ABD=\angle DAB=x$
Now we know that the angle formed on the straight line is ${{180}^{0}}$. So to calculate the angle $\angle ADB$ we will subtract $\angle ADC$ from ${{180}^{0}}$.
$\begin{align}
  & \angle ADB={{180}^{0}}-{{60}^{0}} \\
 & ={{120}^{0}}
\end{align}$

We will now find the value of angle $x$ using the angle sum property of the triangle.
$\begin{align}
  & \angle BAD+\angle ADB+\angle ABD={{180}^{0}} \\
 & x+x+{{120}^{0}}={{180}^{0}} \\
 & 2x={{60}^{0}} \\
 & x={{30}^{0}}
\end{align}$
We will now apply the Law of sine by equating to some constant $k$,
$\begin{align}
  & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\
 & \frac{a}{\sin ({{60}^{0}}+{{30}^{0}})}=\frac{b}{\sin {{30}^{0}}}=\frac{c}{\sin {{60}^{0}}}=k \\
 & \frac{a}{1}=\frac{b}{\frac{1}{2}}=\frac{c}{\frac{\sqrt{3}}{2}}=k \\
 & a=2b=\frac{2c}{\sqrt{3}}=k
\end{align}$
The values of $a,b$and $c$ will be,
$\begin{align}
  & a=k \\
 & b=\frac{k}{2} \\
 & c=\frac{\sqrt{3}k}{2}
\end{align}$
Hence the value of ${{a}^{2}}:{{b}^{2}}:{{c}^{2}}$_{will be,
\[\ {{a}^{2}}:{{b}^{2}}:{{c}^{2}}=1:\frac{1}{4}:\frac{3}{4}\]
\[\ {{a}^{2}}:{{b}^{2}}:{{c}^{2}}=4:1:3\]}
The triangle $\vartriangle ABC$ on which $D$ is the middle point on the side $BC$of a triangle forming an equilateral triangle$\vartriangle ADC$_{, then ${{a}^{2}}:{{b}^{2}}:{{c}^{2}}$ is equal to $4:1:3$}.Hence the correct option is (B).
Note:
The angle sum property which we have used in the solution to find the value of $x$, is a property of a triangle which states that the sum of all the interior angles in a triangle is equal to {{180}^{0}}. This property applies on each and every type of triangle whether it is acute, obtuse or a right angled triangle.