Question

In an acute-angled triangle, express a median in terms of its sides.

Answer 1

Hint: To solve this question we will consider a \[\Delta ABC\] which will have median from point A to side BC and then we will construct a perpendicular from point A to side BC intersecting BC at E. Then, by using Pythagoras Theorem for a different right-angled triangle, we will find the required expression.

Complete step-by-step answer:
In this question, we have to prove that for an acute-angled triangle, we can express the median in terms of its sides. Let us consider an acute-angled triangle ABC which has a median from point A to side BC and name it AD.

Now, we will construct an altitude on the side BC from point A intersecting BC at E. So, we will now use Pythagoras Theorem to find the required expression. We are applying Pythagoras Theorem in \[\Delta AED\]. So, we can write,
\[A{{E}^{2}}+E{{D}^{2}}=A{{D}^{2}}\]
\[A{{E}^{2}}=A{{D}^{2}}-E{{D}^{2}}....\left( i \right)\]
Now, in \[\Delta AEB\], we will apply the Pythagoras theorem. So, we will get,
\[A{{B}^{2}}=A{{E}^{2}}+B{{E}^{2}}....\left( ii \right)\]
Now, we will put the value of \[{{\left( AE \right)}^{2}}\] from equation (i) to equation (ii), we will get,
\[{{\left( AB \right)}^{2}}={{\left( AD \right)}^{2}}-{{\left( ED \right)}^{2}}+{{\left( BE \right)}^{2}}\]
Now from the figure, we can see that BE = BD + DE. So, we can write the equation as,
\[{{\left( AB \right)}^{2}}={{\left( AD \right)}^{2}}-{{\left( DE \right)}^{2}}+{{\left( BD+DE \right)}^{2}}\]
Now, we will expand \[{{\left( BD+DE \right)}^{2}}\] by using the formula
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
So, we will get,
\[{{\left( AB \right)}^{2}}={{\left( AD \right)}^{2}}-{{\left( DE \right)}^{2}}+{{\left( BD \right)}^{2}}+{{\left( DE \right)}^{2}}+2\left( BD \right)\left( DE \right)\]
\[{{\left( AB \right)}^{2}}={{\left( AD \right)}^{2}}+{{\left( BD \right)}^{2}}+2\left( BD \right)\left( DE \right)\]
Now, we can see that AD bisects BC. So, we can say that
\[BD=DC=\dfrac{BC}{2}....\left( iii \right)\]
Therefore, we can write the equation as,
\[{{\left( AB \right)}^{2}}={{\left( AD \right)}^{2}}+{{\left( \dfrac{BC}{2} \right)}^{2}}+2.\left( \dfrac{BC}{2} \right)\left( DE \right)\]
\[{{\left( AB \right)}^{2}}={{\left( AD \right)}^{2}}+\dfrac{{{\left( BC \right)}^{2}}}{4}+\left( BC \right)\left( DE \right)....\left( iv \right)\]
Now, we will apply Pythagoras Theorem in \[\Delta AEC\]. So, we will get,
\[A{{C}^{2}}=A{{E}^{2}}+E{{C}^{2}}\]
Now, we will put the value of \[A{{E}^{2}}\] from equation (i). So, we will get,
\[{{\left( AC \right)}^{2}}={{\left( AD \right)}^{2}}-{{\left( ED \right)}^{2}}+{{\left( EC \right)}^{2}}\]
From the triangle, we can see that EC = CD – DE. So, we can write the equation as,
\[{{\left( AC \right)}^{2}}={{\left( AD \right)}^{2}}-{{\left( ED \right)}^{2}}+{{\left( CD-DE \right)}^{2}}\]
Now, we will expand \[{{\left( CD-DE \right)}^{2}}\] by using the formula \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. So, we will get,
\[{{\left( AC \right)}^{2}}={{\left( AD \right)}^{2}}-{{\left( ED \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( ED \right)}^{2}}-2\left( CD \right)\left( ED \right)\]
\[{{\left( AC \right)}^{2}}={{\left( AD \right)}^{2}}+{{\left( CD \right)}^{2}}-2\left( CD \right)\left( DE \right)\]
From equation (iii), we will put the value of the CD. So, we will get,
\[{{\left( AC \right)}^{2}}={{\left( AD \right)}^{2}}+{{\left( \dfrac{BC}{2} \right)}^{2}}-2\left( \dfrac{BC}{2} \right)\left( DE \right)\]
\[{{\left( AC \right)}^{2}}={{\left( AD \right)}^{2}}+\dfrac{{{\left( BC \right)}^{2}}}{4}-\left( BC \right)\left( DE \right)....\left( v \right)\]
Now, we will add the equation (iv) and (v). So, we will get,
\[{{\left( AB \right)}^{2}}+{{\left( AC \right)}^{2}}={{\left( AD \right)}^{2}}+{{\left( AD \right)}^{2}}+\dfrac{{{\left( BC \right)}^{2}}}{4}+\dfrac{{{\left( BC \right)}^{2}}}{4}\]
\[A{{B}^{2}}+A{{C}^{2}}=2A{{D}^{2}}+\dfrac{B{{C}^{2}}}{2}\]
\[2A{{B}^{2}}+2A{{C}^{2}}=4A{{D}^{2}}+B{{C}^{2}}\]
\[4A{{D}^{2}}=2A{{B}^{2}}+2A{{C}^{2}}-B{{C}^{2}}\]
Hence, we have expressed the median AD in terms of side AB, AC and BC as
\[4A{{D}^{2}}=2A{{B}^{2}}+2A{{C}^{2}}-B{{C}^{2}}\]
Note: We have to remember this relation because if we will get this question in the exam, then we can verify the solution by checking the result because under pressure, we might make a lot of mistakes.