Question

Text for question can start here.
This can have multiple lines. In fig., AD is the median of the triangle ABC and AM \[\bot \]BC. Prove that
(1) \[A{{C}^{2}}=A{{D}^{2}}+BC.DM+{{\left( \dfrac{BC}{2} \right)}^{2}}\]
(2) \[A{{B}^{2}}=A{{D}^{2}}-BC.DM+{{\left( \dfrac{BC}{2} \right)}^{2}}\]
(3) \[A{{B}^{2}}+A{{C}^{2}}=2A{{D}^{2}}+\dfrac{B{{C}^{2}}}{2}\]

Answer 1

Hint: Consider MD = x. Then BM=BC/2-x and use Pythagoras theorem to the right-angled triangles AMB and AMC. Since, AD is the median\[\Rightarrow \] BD = \[\dfrac{BC}{2}\], CD = \[\dfrac{BC}{2}\]

Complete step-by-step answer:

We are given the triangle ABC in which AD is the median and AM \[\bot \]BC . Since, AD is the median, BD = \[\dfrac{BC}{2}\] and CD = \[\dfrac{BC}{2}\] .
Now, let’s assume MD = x units.
\[\Rightarrow BM=\dfrac{BC}{2}-x\] and $CM=\dfrac{BC}{2}+x$
Now, we are given that AM \[\bot \]BC. So, $\angle AMB={{90}^{o}}$ . So, $\Delta AMB$ is a right-angled triangle. So, we can apply Pythagoras theorem in $\Delta AMB$ . But before applying Pythagoras theorem, we should understand what Pythagoras theorem is. The Pythagoras theorem states that “In a right-angled triangle, the sum of squares of the perpendicular sides is equal to the square of the third side, or the hypotenuse.”
So, in $\Delta AMB$ ,
\[A{{M}^{2}}+B{{M}^{2}}=A{{B}^{2}}\]
\[\Rightarrow A{{M}^{2}}+{{\left( \dfrac{BC}{2}-x \right)}^{2}}=A{{B}^{2}}\] \[......(1)\]
Now, in $\Delta AMC$ , $\angle AMC={{90}^{o}}$. So, $\Delta AMC$ is a right-angled triangle. So, we can apply Pythagoras theorem in $\Delta AMC$ .
So, in $\Delta AMC$ ,
\[A{{M}^{2}}+C{{M}^{2}}=A{{B}^{2}}\]
\[\Rightarrow A{{M}^{2}}+{{\left( \dfrac{BC}{2}+x \right)}^{2}}=A{{C}^{2}}\] \[......(2)\]
Now, we will subtract equation 1 from equation 2. So, we get:
\[{{\left( \dfrac{BC}{2}+x \right)}^{2}}-{{\left( \dfrac{BC}{2}-x \right)}^{2}}=A{{C}^{2}}-A{{B}^{2}}\]
$\Rightarrow \dfrac{B{{C}^{2}}}{4}+BC\times x+{{x}^{2}}-\left( \dfrac{B{{C}^{2}}}{4}-BC\times x+{{x}^{2}} \right)=A{{C}^{2}}-A{{B}^{2}}$
$\Rightarrow \dfrac{B{{C}^{2}}}{4}+BC\times x+{{x}^{2}}-\dfrac{B{{C}^{2}}}{4}+BC\times x-{{x}^{2}}=A{{C}^{2}}-A{{B}^{2}}$
$\Rightarrow 2BC\times x=A{{C}^{2}}-A{{B}^{2}}$
We had assumed DM = x.
\[\Rightarrow 2.\left( BC \right).DM+A{{B}^{2}}=A{{C}^{2}}\]\[...(3)\]
Now, consider the triangle AMD . We have $\angle AMD={{90}^{o}}$ . So, $\Delta AMD$ is a right-angled triangle. So, we can apply Pythagoras theorem. On applying Pythagoras theorem, we get:
\[A{{M}^{2}}+{{x}^{2}}=A{{D}^{2}}\]
From equation 2:
\[A{{M}^{2}}+{{\left( \dfrac{BC}{2}+x \right)}^{2}}=A{{C}^{2}}\]
\[\Rightarrow A{{D}^{2}}-{{x}^{2}}+{{\left( \dfrac{BC}{2}+x \right)}^{2}}=A{{C}^{2}}\]
\[\Rightarrow A{{C}^{2}}=A{{D}^{2}}+BC.x+{{\left( \dfrac{BC}{2} \right)}^{2}}\]
\[\Rightarrow A{{C}^{2}}=A{{D}^{2}}+BC.DM+{{\left( \dfrac{BC}{2} \right)}^{2}}........(4)\]
Hence, (i) is proved.
Now, from equation 3:
\[A{{C}^{2}}=A{{D}^{^{2}}}+\dfrac{A{{C}^{2}}-A{{B}^{2}}}{2}+{{\left( \dfrac{BC}{2} \right)}^{2}}\]
\[\Rightarrow A{{C}^{2}}-\dfrac{A{{C}^{2}}-A{{B}^{2}}}{2}=A{{D}^{^{2}}}+{{\left( \dfrac{BC}{2} \right)}^{2}}\]
\[\Rightarrow \dfrac{A{{C}^{2}}+A{{B}^{2}}}{2}=A{{D}^{^{2}}}+{{\left( \dfrac{BC}{2} \right)}^{2}}\]
\[\Rightarrow A{{C}^{2}}+A{{B}^{2}}=2A{{D}^{^{2}}}+\dfrac{B{{C}^{2}}}{2}........(5)\]
Hence, (iii) is proved.
On subtracting equation (4) from equation (5), we get:
\[A{{C}^{2}}+A{{B}^{2}}-A{{C}^{2}}=2A{{D}^{^{2}}}+\dfrac{B{{C}^{2}}}{2}-\left( A{{D}^{2}}+BC.DM+{{\left( \dfrac{BC}{2} \right)}^{2}} \right)\]
$\Rightarrow A{{B}^{2}}=2A{{D}^{^{2}}}+\dfrac{B{{C}^{2}}}{2}-A{{D}^{2}}-BC.DM-{{\left( \dfrac{BC}{2} \right)}^{2}}$
\[\Rightarrow \]\[A{{B}^{2}}=A{{D}^{2}}-BC.DM+{{\left( \dfrac{BC}{2} \right)}^{2}}\]
Hence, (ii) is proved.

NOTE: While calculating, be careful about the signs. Sign mistakes are very common and can result in wrong answers. So, students should be very careful while calculating and while making substitutions.