Question
Prove that the medians of an equilateral triangle are equal.
✓
Answer
Given,
To prove the medians of an equilateral triangle are equal.
Median: The line Joining the vertex and midpoint of opposite side. Now, consider an equilateral triangle ABC.
Let D, E, F are midpoints of BC, CA and AB.
Then, AD, BE and CF are medians of ABC.
Now,
D is midpoint of $\text{BC}\Rightarrow\text{BD}=\text{DC}=\frac{\text{BC}}{2}$
Similarly, $\text{CE}=\text{EA}=\frac{\text{AC}}{2}$
$\text{AF}=\text{FB}=\frac{\text{AB}}{2}$
Since $\triangle\text{ABC}$ is an equilateral triangle
$\Rightarrow\text{AB}=\text{BC}=\text{CA}\ ...(\text{i)}$
$\Rightarrow\text{BD}=\text{DC}=\text{CE}=\text{EA}=\text{AF}=\text{FB}\\=\frac{\text{BC}}{2}=\frac{\text{AC}}{2}=\frac{\text{AB}}{2}\ ...(\text{ii})$
And also, $\angle\text{ABC}=\angle\text{BCA}=\angle\text{CAB}=60^\circ\ ....(\text{iii)}$
Now, consider $\triangle\text{ABC}$ and $\triangle\text{BCE}\text{ AB}=\text{BC}$ [From (i)]
$\text{BD}=\text{CE}$ [From (ii)]
Now, in $\triangle\text{TSR}$ and $\triangle\text{TRQ}$
$\text{TS}=\text{TR}$ [From (iii)]
$\angle\text{ABD}=\angle\text{BCE}$ [From (iii)] $\big[\angle\text{ABD}$ and $\angle\text{ABC}$ and $\angle\text{BCE}$ and $\angle\text{BCA}$ are same $\big]$
So, from SAS congruence criterion, we have
$\triangle\text{ABD}=\triangle\text{BCE}$
$\text{AD}=\text{BE}\ ...(\text{iv})$
[Corresponding parts of congruent triangles are equal]
Now, consider $\triangle\text{BCE}$ and $\triangle\text{CAF},\text{BC}=\text{CA}$ [From (i)]
$\angle\text{BCE}=\angle\text{CAF}$ [From (ii)]
$\big[\angle\text{BCE}$ and $\angle\text{BCA}$ and $\angle\text{CAF}$ and $\angle\text{CAB}$ are same $\big]$
$\text{CE}=\text{AF}$ [From (ii)]
So, from SAS congruence criterion, we have
$\triangle\text{BCE}=\triangle\text{CAF}$
[Corresponding parts of congruent triangles are equal]
From (iv) and (v), we have
AD = BE = CF
Median AD = Median BE = Median CF
The medians of an equilateral triangle are equal.
Hence proved
To prove the medians of an equilateral triangle are equal.
Median: The line Joining the vertex and midpoint of opposite side. Now, consider an equilateral triangle ABC.
Let D, E, F are midpoints of BC, CA and AB.
Then, AD, BE and CF are medians of ABC.
Now,
D is midpoint of $\text{BC}\Rightarrow\text{BD}=\text{DC}=\frac{\text{BC}}{2}$
Similarly, $\text{CE}=\text{EA}=\frac{\text{AC}}{2}$
$\text{AF}=\text{FB}=\frac{\text{AB}}{2}$
Since $\triangle\text{ABC}$ is an equilateral triangle
$\Rightarrow\text{AB}=\text{BC}=\text{CA}\ ...(\text{i)}$
$\Rightarrow\text{BD}=\text{DC}=\text{CE}=\text{EA}=\text{AF}=\text{FB}\\=\frac{\text{BC}}{2}=\frac{\text{AC}}{2}=\frac{\text{AB}}{2}\ ...(\text{ii})$
And also, $\angle\text{ABC}=\angle\text{BCA}=\angle\text{CAB}=60^\circ\ ....(\text{iii)}$
Now, consider $\triangle\text{ABC}$ and $\triangle\text{BCE}\text{ AB}=\text{BC}$ [From (i)]
$\text{BD}=\text{CE}$ [From (ii)]
Now, in $\triangle\text{TSR}$ and $\triangle\text{TRQ}$
$\text{TS}=\text{TR}$ [From (iii)]
$\angle\text{ABD}=\angle\text{BCE}$ [From (iii)] $\big[\angle\text{ABD}$ and $\angle\text{ABC}$ and $\angle\text{BCE}$ and $\angle\text{BCA}$ are same $\big]$
So, from SAS congruence criterion, we have
$\triangle\text{ABD}=\triangle\text{BCE}$
$\text{AD}=\text{BE}\ ...(\text{iv})$
[Corresponding parts of congruent triangles are equal]
Now, consider $\triangle\text{BCE}$ and $\triangle\text{CAF},\text{BC}=\text{CA}$ [From (i)]
$\angle\text{BCE}=\angle\text{CAF}$ [From (ii)]
$\big[\angle\text{BCE}$ and $\angle\text{BCA}$ and $\angle\text{CAF}$ and $\angle\text{CAB}$ are same $\big]$
$\text{CE}=\text{AF}$ [From (ii)]
So, from SAS congruence criterion, we have
$\triangle\text{BCE}=\triangle\text{CAF}$
[Corresponding parts of congruent triangles are equal]
From (iv) and (v), we have
AD = BE = CF
Median AD = Median BE = Median CF
The medians of an equilateral triangle are equal.
Hence proved
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