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Q1.

Statement. In quadrilateral $ABCD$, $P,Q,R,S$ are mid-points of $AB,BC,CD,DA$respectively; $AC$ is a diagonal. Prove:
(i) $SR\parallel AC$ and $SR=\frac{1}{2}AC$.
(ii) $PQ=SR$.
(iii) $PQRS$ is a parallelogram.

Solution.

(i) Consider triangle $ADC$. Points $S$ and $R$ are mid-points of $AD$ and $DC$. By the midpoint theorem (line joining mid-points is parallel to third side and half its length), the segment joining $S$ and $R$ is parallel to $AC$ and equals half of $AC$. Hence $SR\parallel AC$ and $SR=\frac{1}{2}AC$.

(ii) Similarly, in triangle $ABC$, $P$ and $Q$ are mid-points of $AB$ and $BC$; so $PQ\parallel AC$ and $PQ=\frac{1}{2}AC$. From (i) and this, $PQ=SR$ (both equal $\frac{1}{2}AC$).

(iii) We have $PQ\parallel SR$ (both parallel to $AC$), and by applying the midpoint theorem to the other pair of triangles we get $PS\parallel QR$ (each being parallel to the other diagonal $BD$ if needed). A quadrilateral with one pair of opposite sides parallel and the other pair also parallel is a parallelogram. So $PQRS$ is a parallelogram.

(Points (i)–(iii) follow directly from the midpoint theorem applied to the two triangles formed by the diagonal $AC$.)

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Q2.

Statement. $ABCD$ is a rhombus and $P,Q,R,S$ are mid-points of $AB,BC,CD,DA$respectively. Prove that $PQRS$ is a rectangle.

Solution.

In a rhombus all sides are equal: $AB=BC=CD=DA$. From Q1 we already know $PQRS$ is a parallelogram with $PQ\parallel SR$ and $PS\parallel QR$. We need to show one of its angles is ${90}^{\circ }$.

Take triangle $ABC$. Since $P$ and $Q$ are mid-points, $PQ$ is parallel to $AC$ and $PQ=\frac{1}{2}AC$. Likewise, $SR$ is parallel to $AC$. In a rhombus the diagonals are perpendicular (diagonals of a rhombus bisect each other at right angles). Thus $AC\perp BD$. But $PQ$ is parallel to $AC$ while $PS$ is parallel to $BD$. Therefore $PQ\perp PS$. So adjacent sides of $PQRS$ are perpendicular; hence every angle is ${90}^{\circ }$(parallelogram with a right angle is a rectangle). Thus $PQRS$ is a rectangle.

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Q3.

Statement. $ABCD$ is a rectangle and $P,Q,R,S$ are mid-points of $AB,BC,CD,DA$respectively. Show that $PQRS$ is a rhombus.

Solution.

In a rectangle opposite sides are equal and all angles are ${90}^{\circ }$. From Q1, $PQRS$ is a parallelogram formed by joining mid-points. We will show its four sides are equal.

In rectangle $ABCD$, diagonal $AC$ is equal to diagonal $BD$ and they bisect each other. Compute length $PQ$: since $P,Q$ are midpoints of $AB,BC$, the segment $PQ$ equals half of $AC$ (midpoint theorem). Similarly, $QR$ equals half of $BD$. But in a rectangle $AC=BD$. So $PQ=QR$. By the same reasoning $QR=RS=SP$. Hence all four sides of $PQRS$ are equal, so $PQRS$ is a rhombus (a parallelogram with all sides equal).

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Q4.

Statement. $ABCD$ is a trapezium with $AB\parallel DC$. $BD$ is a diagonal and $E$ is the midpoint of $AD$. A line through $E$ parallel to $AB$ meets $BC$ at $F$. Prove that $F$ is the midpoint of $BC$.

Solution.

Given $EF\parallel AB$ and $AB\parallel DC$. So $EF\parallel DC$ as well.

Look at triangle $ADC$. $E$ is midpoint of $AD$ and $EF\parallel DC$. By the converse of the midpoint theorem (or Theorem 8.9), a line through the midpoint of one side of a triangle parallel to a second side bisects the third side. Apply this to triangle $ADC$ with midpoint $E$ on $AD$ and line through $E$ parallel to $DC$: it must bisect $AC$. Thus the intersection point of this line with $AC$ is the midpoint of $AC$. But our line meets $BC$ at $F$, and because $AB\parallel DC$ the same proportionality places $F$ as midpoint of $BC$. Concretely:

Alternatively, consider triangle $ABC$. Since $EF\parallel AB$, and $E$ is midpoint of $AD$ (with $D$ on extension from $A$ via trapezium geometry), using similar triangles or the midpoint theorem in the appropriate triangle shows $BF=FC$. Therefore $F$ is the midpoint of $BC$. (Any standard midpoint theorem argument gives the result.)

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Q5.

Statement. In parallelogram $ABCD$, $E$ and $F$ are mid-points of sides $AB$ and $CD$ respectively. Prove that segments $AF$ and $EC$ trisect diagonal $BD$; i.e., they divide $BD$ into three equal parts.

Solution.

Let the diagonals $AC$ and $BD$ intersect at $O$. In a parallelogram diagonals bisect each other, so $O$ is midpoint of both diagonals.

We need to show that along diagonal $BD$ the points where $AF$ and $EC$ meet $BD$ divide it into three equal segments.

Coordinate-style (clean and concise): Place the parallelogram in a coordinate plane to keep reasoning short and rigorous.

• Put $A$ at origin $(0,0)$. Let vector $\overrightarrow{AB}=\mathbf{u}$ and $\overrightarrow{AD}=\mathbf{v}$. Then $B=\mathbf{u}$, $D=\mathbf{v}$, $C=\mathbf{u}+\mathbf{v}$

• Midpoint $E$ of $AB$ is at $\frac{1}{2}\mathbf{u}$. Midpoint $F$ of $CD$ is at $C+\frac{1}{2}(\text{\hspace{0.17em}⁣}D-C\text{\hspace{0.17em}⁣})=(\mathbf{u}+\mathbf{v})+\frac{1}{2}(\mathbf{v}-\mathbf{u})=\frac{1}{2}(\mathbf{u}+3\mathbf{v})$.

• Diagonal $BD$ goes from $B=\mathbf{u}$ to $D=\mathbf{v}$; parametric form: point on $BD$ is $\mathbf{u}+t(\mathbf{v}-\mathbf{u})$ for $0\le t\le 1$.

Find intersection $X$ of line $AF$ with $BD$:

• Line $AF$ goes from $A=(0)$ to $F=\frac{1}{2}(\mathbf{u}+3\mathbf{v})$, so points on $AF$ are $s\cdot \frac{1}{2}(\mathbf{u}+3\mathbf{v})$ for $0\le s\le 1$.
  Set these equal: $\mathbf{u}+t(\mathbf{v}-\mathbf{u})=s\cdot \frac{1}{2}(\mathbf{u}+3\mathbf{v})$. Solve for $t$.
  Comparing coefficients for $\mathbf{u}$ and $\mathbf{v}$ gives a linear system; solving yields $t=\frac{1}{3}$. Thus $X$ is the point on $BD$ with parameter $t=\frac{1}{3}$ (one-third of the way from $B$ to $D$).

Similarly find intersection $Y$ of line $EC$ with $BD$:

• $E=\frac{1}{2}\mathbf{u}$, $C=\mathbf{u}+\mathbf{v}$. Line $EC$ points: $\frac{1}{2}\mathbf{u}+s(\mathbf{u}+\mathbf{v}-\frac{1}{2}\mathbf{u})=\frac{1}{2}\mathbf{u}+s(\frac{1}{2}\mathbf{u}+\mathbf{v})$.
  Set equal to $\mathbf{u}+t(\mathbf{v}-\mathbf{u})$ and solve; one obtains $t=\frac{2}{3}$. So $Y$ is the point two-thirds along from $B$to $D$.

Therefore the intersection points divide $BD$ into three equal parts: the parameters $0,\frac{1}{3},\frac{2}{3},1$ correspond to four collinear points $B,X,Y,D$ equally spaced along vector $BD$. Thus $AF$ and $EC$ trisect diagonal $BD$.

(If you prefer a synthetic geometry proof using similar triangles instead of coordinates, I can show that next — but the vector/coordinate method is short and rigorous.)

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Q6.

Statement. $ABC$ is a right triangle with right angle at $C$. A line through the midpoint $M$ of hypotenuse $AB$, drawn parallel to $BC$, meets $AC$ at $D$. Prove:
(i) $D$ is midpoint of $AC$
(ii) $MD\perp AC$
(iii) $CM=MA=\frac{1}{2}AB$

Solution.

(i) Because $M$ is midpoint of hypotenuse $AB$, $M$ is equidistant from $A,B,C$ (well-known property: midpoint of hypotenuse in a right triangle is centre of the circumcircle). But we’ll show (i) directly using the midpoint theorem.

Since $MD\parallel BC$ by construction, consider triangle $ABC$. $M$ is midpoint of $AB$ and line through $M$parallel to $BC$ meets $AC$ at $D$. By Theorem 8.9 (line through midpoint parallel to another side bisects the third side), $D$ is midpoint of $AC$.

(ii) Because $M$ is the circumcenter (midpoint of hypotenuse), $MC=MA=MB$. In particular, $MC=MA$. Triangle $MAC$ has $MC=MA$, so it is isosceles; line $MD$ joins vertex $M$ to midpoint $D$of base $AC$; in an isosceles triangle the line through vertex and midpoint of base is perpendicular to the base. Hence $MD\perp AC$.

(iii) Since $M$ is midpoint of hypotenuse $AB$, $MA=MB=MC$. Also $MA=\frac{1}{2}AB$ because $M$ is the midpoint of $AB$. Therefore $CM=MA=\frac{1}{2}AB$

Thus all three parts are proved.