Courses
Courses for Kids
Free study material
Offline Centres
More

# Triangle ABC is right angled at B and D is the midpoint of side BC. Prove that $A{C^2} = 4A{D^2} - 3A{B^2}$.

Last updated date: 26th Feb 2024
Total views: 359.7k
Views today: 11.59k
Verified
359.7k+ views
Hint:
Here, we will use the given information to write BD in terms of BC. Then, we will use Pythagoras’s theorem in both triangle ADB and ACB. We will simplify the equations obtained using the value of BD, and then compare the two equations to prove that $A{C^2} = 4A{D^2} - 3A{B^2}$.

Formula Used:
We will use the formula of the Pythagoras’s theorem, ${\rm{Hypotenuse}}{{\rm{e}}^2} = {\rm{Bas}}{{\rm{e}}^2} + {\rm{Perpendicula}}{{\rm{r}}^2}$.

Complete step by step solution:
First, we will draw the figure of the given triangle.

Here, D is the midpoint of BC and $\angle B = 90^\circ$.
Since D is the midpoint of BC, we get
$BD = DC$
We can observe that $BD + DC = BC$.
Substituting $BD = DC$ in the equation, we get
$\Rightarrow BD + BD = BC$
$\Rightarrow 2BD = BC$
Dividing both sides by 2, we get
$\Rightarrow BD = \dfrac{1}{2}BC$
Therefore, we get
$\Rightarrow BD = \dfrac{1}{2}BC$
Next, we will use the Pythagoras’s theorem.
The Pythagoras’s theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of squares of the other two sides, that is ${\rm{Hypotenuse}}{{\rm{e}}^2} = {\rm{Bas}}{{\rm{e}}^2} + {\rm{Perpendicula}}{{\rm{r}}^2}$.
Therefore, in triangle ADB, we get
$A{D^2} = B{D^2} + A{B^2}$
Substituting $BD = \dfrac{1}{2}BC$ in the equation, we get
$\Rightarrow A{D^2} = {\left( {\dfrac{1}{2}BC} \right)^2} + A{B^2}$
Simplifying the expression, we get
$\Rightarrow A{D^2} = \dfrac{1}{4}B{C^2} + A{B^2}$
Subtracting $A{B^2}$ from both sides of the equation, we get
$\begin{array}{l} \Rightarrow A{D^2} - A{B^2} = \dfrac{1}{4}B{C^2} + A{B^2} - A{B^2}\\ \Rightarrow A{D^2} - A{B^2} = \dfrac{1}{4}B{C^2}\end{array}$
Multiplying both sides of the equation by 4, we get
$\Rightarrow 4\left( {A{D^2} - A{B^2}} \right) = \dfrac{1}{4}B{C^2} \times 4$
Simplifying the expression using the distributive law of multiplication, we get
$\Rightarrow 4A{D^2} - 4A{B^2} = B{C^2}$
Now, using the Pythagoras’s theorem in triangle ACB, we get
$A{C^2} = B{C^2} + A{B^2}$
Subtracting $A{B^2}$ from both sides, we get
$\begin{array}{l} \Rightarrow A{C^2} - A{B^2} = B{C^2} + A{B^2} - A{B^2}\\ \Rightarrow A{C^2} - A{B^2} = B{C^2}\end{array}$
From the equations $A{C^2} - A{B^2} = B{C^2}$ and $4A{D^2} - 4A{B^2} = B{C^2}$, we get
$\Rightarrow A{C^2} - A{B^2} = 4A{D^2} - 4A{B^2}$
Adding $A{B^2}$ to both sides, we get
$\Rightarrow A{C^2} - A{B^2} + A{B^2} = 4A{D^2} - 4A{B^2} + A{B^2}$
Therefore, we get
$\Rightarrow A{C^2} = 4A{D^2} - 3A{B^2}$

Hence, we have proved that $A{C^2} = 4A{D^2} - 3A{B^2}$.

Note:
We have used the distributive law of multiplication to multiply 4 by $\left( {A{D^2} - A{B^2}} \right)$. The distributive law of multiplication states that $a\left( {b + c} \right) = a \cdot b + a \cdot c$.
The point D is the midpoint of side BC. This means that AD bisects BC into two equal parts. Therefore, AD is a median of the given triangle ABC. A triangle is a two dimensional closed geometric shape which has 3 sides.