NCERT Solutions

it bisects $\angle \mathrm{C}$ also,

ABCD is a rhombus.

APCQ is a parallelogram

diagonal $\mathrm{AC}=$ diagonal BD [Hint: Extend AB and draw a line through C

parallel to DA intersecting AB produced at E .]

\subsection*{8.2 The Mid-point Theorem}

You have studied many properties of a triangle as well as a quadrilateral. Now let us study yet another result which is related to the mid-point of sides of a triangle. Perform the following activity.

Draw a triangle and mark the mid-points E and F of two sides of the triangle. Join the points E and F (see Fig. 8.15).

Measure EF and BC . Measure $\angle \mathrm{AEF}$ and $\angle \mathrm{ABC}$. What do you observe? You will find that : $\mathrm{EF}=\frac{1}{2} \mathrm{BC} \text { and } \angle \mathrm{AEF}=\angle \mathrm{ABC}$ so, $\mathrm{EF} \| \mathrm{BC}$ Repeat this activity with some more triangles. So, you arrive at the following theorem:

Theorem 8.8: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

You can prove this theorem using the following clue:

Observe Fig 8.16 in which E and F are mid-points of AB and AC respectively and $\mathrm{CD} \| \mathrm{BA}$. $\Delta \mathrm{AEF} \cong \Delta \mathrm{CDF} \quad \text { (ASA Rule) }$

So, $\quad \mathrm{EF}=\mathrm{DF}$ and $\mathrm{BE}=\mathrm{AE}=\mathrm{DC} \quad$ (Why?) Therefore, BCDE is a parallelogram. (Why?) This gives $\mathrm{EF} \| \mathrm{BC}$.

In this case, also note that $\mathrm{EF}=\frac{1}{2} \mathrm{ED}=\frac{1}{2} \mathrm{BC}$. Can you state the converse of Theorem 8.8? Is the converse true? You will see that converse of the above theorem is also true which is stated as below:

Theorem 8.9 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

In Fig 8.17, observe that E is the mid-point of AB , line $l$ is passsing through E and is parallel to BC and $\mathrm{CM} \| \mathrm{BA}$.

Prove that $\mathrm{AF}=\mathrm{CF}$ by using the congruence of $\triangle \mathrm{AEF}$ and $\triangle \mathrm{CDF}$.

Example 6 : In $\triangle \mathrm{ABC}, \mathrm{D}, \mathrm{E}$ and F are respectively the mid-points of sides $\mathrm{AB}, \mathrm{BC}$ and CA (see Fig. 8.18). Show that $\triangle \mathrm{ABC}$ is divided into four congruent triangles by joining $\mathrm{D}, \mathrm{E}$ and F . Solution : As D and E are mid-points of sides AB and BC of the triangle ABC , by Theorem 8.8,

\section*{$\mathrm{DE} \| \mathrm{AC}$}

Similarly, $\mathrm{DF} \| \mathrm{BC}$ and $\mathrm{EF} \| \mathrm{AB}$

Therefore $\mathrm{ADEF}, \mathrm{BDFE}$ and DFCE are all parallelograms. Now DE is a diagonal of the parallelogram BDFE , therefore, $\quad \triangle \mathrm{BDE} \cong \triangle \mathrm{FED}$ Similarly $\quad \triangle \mathrm{DAF} \cong \triangle \mathrm{FED}$ and $\quad \triangle \mathrm{EFC} \cong \triangle \mathrm{FED}$ So, all the four triangles are congruent. Example 7 : $l, m$ and $n$ are three parallel lines intersected by transversals $p$ and $q$ such that $l, m$ and $n$ cut off equal intercepts AB and BC on $p$ (see Fig. 8.19). Show that $l, m$ and $n$ cut off equal intercepts DE and EF on $q$ also. Solution : We are given that $\mathrm{AB}=\mathrm{BC}$ and have to prove that $\mathrm{DE}=\mathrm{EF}$. Let us join A to F intersecting $m$ at G .. The trapezium ACFD is divided into two triangles;

namely $\triangle \mathrm{ACF}$ and $\triangle \mathrm{AFD}$. In $\triangle \mathrm{ACF}$, it is given that B is the mid-point of $\mathrm{AC}(\mathrm{AB}=\mathrm{BC})$ and $\quad \mathrm{BG} \| \mathrm{CF} \quad($ since $m \| n)$. So, G is the mid-point of AF (by using Theorem 8.9) Now, in $\triangle \mathrm{AFD}$, we can apply the same argument as G is the mid-point of AF , $\mathrm{GE} \| \mathrm{AD}$ and so by Theorem 8.9, E is the mid-point of DF , i.e., $\quad \mathrm{DE}=\mathrm{EF}$. In other words, $l, m$ and $n$ cut off equal intercepts on $q$ also.

$\mathrm{SR} \| \mathrm{AC}$ and $\mathrm{SR}=\frac{1}{2} \mathrm{AC}$

PQRS is a parallelogram.

opposite sides are equal

opposite angles are equal

diagonals bisect each other

D is the mid-point of AC

\subsection*{8.3 Summary}

In this chapter, you have studied the following points :

it bisects $\angle \mathrm{C}$ also,

ABCD is a rhombus.

APCQ is a parallelogram

diagonal $\mathrm{AC}=$ diagonal BD [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E .]

$\mathrm{SR} \| \mathrm{AC}$ and $\mathrm{SR}=\frac{1}{2} \mathrm{AC}$

PQRS is a parallelogram.

D is the mid-point of AC