## Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

[question] Question. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. [/question] [solution] Solution: Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD. In ΔABD, S and P are the mid-points of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that $S P \| B D$ and $S P=\frac{1}{2} B D \ldots$(1) Similarly in $\triangl...

## In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

[question] Question. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD. [/question] [solution] Solution: $A B C D$ is a parallelogram. $\therefore \mathrm{AB} \| \mathrm{CD}$ And hence, $A E \|$ FC Again, $A B=C D$ (Opposite sides of parallelogram $A B C D$ ) $\frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{CD}$ $\mathrm{AE}=\mathrm{FC}(\mathrm{E}$ and $\mathrm{F}$ are mid-points o...

## ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid - point of AD.

[question] Question. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid - point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC. [question] [solution] Solution: Let EF intersect DB at G. By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side. In $\triangle \mathrm{ABD}$, $E F \| A B$ and $E$...

## ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

[question] Question. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. [/question] [solution] Solution: Let us join $\mathrm{AC}$ and $\mathrm{BD}$. In $\triangle \mathrm{ABC}$, $P$ and $Q$ are the mid-points of $A B$ and $B C$ respectively. $\therefore P Q \| A C$ and $P Q=\frac{1}{2} A C$ (Mid-point theorem) $\ldots$(1) Similarly in $\triangle \mathrm{ADC}$, $S R \| A C$ and $S R=\frac{1}{2} A C$ (Mi...

## ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

[question] Question. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. [/question] [solution] Solution: In ΔABC, P and Q are the mid-points of sides AB and BC respectively. $\therefore \mathrm{PQ} \| \mathrm{AC}$ and $\mathrm{PQ}=\frac{1}{2} \mathrm{AC}$ (Using mid-point theorem) $\ldots$(1) In $\triangle \mathrm{ADC}$ $R$ and $S$ are the mid-points of $C D$ and $A D$ respectively. $\therefore R S ...

## ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA

[question] Question. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that: (i) $S R \| A C$ and $S R=\frac{1}{2} A C$ (ii) $P Q=S R$ (iii) $\mathrm{PQRS}$ is a parallelogram. [/question] [solution] Solution: (i) In ΔADC, S and R are the mid-points of sides AD and CD respectively. In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is ha...

## ABCD is a trapezium in which AB || CD and AD = BC

[question] Question. ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that (i) $\angle \mathrm{A}=\angle \mathrm{B}$ (ii) $\angle \mathrm{C}=\angle \mathrm{D}$ (iii) $\triangle \mathrm{ABC} \cong \triangle \mathrm{BAD}$ (iv) diagonal $A C=$ diagonal $B D$ [/question] [solution] Solution: Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram. (i) $A D=C E$ (Opposite sides of parall...