CBSE Class 10 Maths Chapter 6 Triangle Ex 6.4 || NCERT Solutions for Class 10 Maths Chapter 6: Triangle (English Medium)

Given: D, E and F are the mid points of sides \( \mathrm{AB}, \mathrm{BC} \) and CA respectively.
Because \( \mathrm{D}, \mathrm{E} \) and F are respectively the mid-points of sides \( \mathrm{AB}, \mathrm{BC} \) and CA of \( \triangle \mathrm{ABC} \),
Midpoint Theorem: The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side.
Therefore, From mid-point theorem,
\( \mathrm{DE} \| \mathrm{AC} \) and \( \mathrm{DE}=\frac{1}{2} \mathrm{AC} \)
DF|| BC and \( \mathrm{DF}=\frac{1}{2} \mathrm{BC} \)
\( \mathrm{EF} \| \mathrm{AB} \) and \( \mathrm{EF}=\frac{1}{2} \mathrm{AB} \)
Now, In \( \triangle \mathrm{BED} \) and \( \triangle \mathrm{BCA} \)
\( \angle \mathrm{BED}=\angle \mathrm{BCA} \) (Corresponding angles)
\( \angle \mathrm{BDE}=\angle \mathrm{BAC} \) (Corresponding angles)
\( \angle \mathrm{EBD}=\angle \mathrm{CBA} \) (Common angles)
Therefore,
If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar (AAA similarity criterion).
\( \triangle \mathrm{BED} \sim \triangle \mathrm{BCA} \) (From the AAA similarity)
Theorem 6.6: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\( \frac{\operatorname{ar}(\triangle B E D)}{\operatorname{ar}(\triangle B C A)}=\left(\frac{D E}{A C}\right)^{2}\)
\(\frac{\operatorname{ar}(\triangle B E D)}{\operatorname{ar}(\triangle B C A)}=\left(\frac{D E}{2 D E}\right)^{2}\)
\(\frac{\operatorname{ar}(\triangle B E D)}{\operatorname{ar}(\triangle B C A)}=\frac{1}{4}\)
\(\operatorname{ar}(\triangle B E D)=\frac{1}{4} \operatorname{ar}(\triangle B C A) \)
Similarly,
\( \operatorname{ar}(\triangle C F E)=\frac{1}{4} \operatorname{ar}(\triangle C B A) \)
And,
\( \operatorname{ar}(\triangle A D F)=\frac{1}{4} \operatorname{ar}(\triangle A B C) \)
Also,
\( \operatorname{ar}(\triangle D E F)=\operatorname{ar}(\triangle A B C)-[\operatorname{ar}(\triangle B E D)+\operatorname{ar}(\triangle C F E)+\operatorname{ar}(\triangle A D F)]\)
\(\operatorname{ar}(\triangle D E F)=\operatorname{ar}(\triangle \mathrm{ABC})-\frac{3}{4} \operatorname{ar}(\triangle A B C)\)
\(=\frac{1}{4} \operatorname{ar}(\triangle A B C)\)
\(\frac{\operatorname{ar}(\triangle D E F)}{\operatorname{ar}(\triangle A B C)}=\frac{1}{4} \)

Let us assume two similar triangles as \( \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \).
Let AD and PS be the medians of these triangles
Then, because \( \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \)
\( \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\quad\ldots \text{(i)}\)
\(\angle \mathrm{A}=\angle \mathrm{P}, \angle \mathrm{B}=\angle \mathrm{Q}, \angle \mathrm{C}=\angle \mathrm{R} \quad\ldots \text{(ii)}\)
Since AD and PS are medians,
\( \mathrm{BD}=\mathrm{DC}= \frac{ \mathrm{BC} }{ {2} } \)
And, \( \mathrm{QS}=\mathrm{SR}= \frac{ \mathrm{QR} }{ {2} } \)
Equation (i) becomes,
\( \frac{A B}{P Q}=\frac{B D}{Q S}=\frac{A C}{P R} \quad\ldots \text{(iii)}\)
In \( \triangle \mathrm{ABD} \) and \( \triangle \mathrm{PQS} \),
\( \angle \mathrm{B}=\angle \mathrm{Q}\quad[ \)From (ii)\( ] \)
And
\( \frac{A B}{P Q}=\frac{B D}{Q S} \quad[ \)From (iii)\( ] \)
\( \triangle \mathrm{ABD} \sim \triangle \mathrm{PQS} \) (SAS similarity)
Therefore, it can be said that
\( \frac{A B}{P Q}=\frac{B D}{Q s}=\frac{A D}{P S} \quad\ldots \text{(iv)}\)
\( \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\left(\frac{\triangle A B}{\triangle P Q}\right)^{2}=\left(\frac{B C}{Q R}\right)^{2}=\left(\frac{A C}{P R}\right)^{2} \)
From (i) and (iv), we get
\( \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}=\frac{A D}{P S} \)
And hence,
\( \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\left(\frac{\triangle A D}{\triangle P S}\right)^{2} \)