Exercise 8.2 Practice
The Mid-point Theorem
Q1: Mid-points Quadrilateral
00:00
$ABCD$ is a quadrilateral in which $P, Q, R$ and $S$ are mid-points of the sides $AB, BC, CD$ and $DA$. $AC$ is a diagonal. Show that:
(i) $SR \parallel AC$ and $SR = \frac{1}{2} AC$
(ii) $PQ = SR$
(iii) $PQRS$ is a parallelogram.
(i) $SR \parallel AC$ and $SR = \frac{1}{2} AC$
(ii) $PQ = SR$
(iii) $PQRS$ is a parallelogram.
(i) In $\triangle ADC$:
$S$ is the mid-point of $DA$ and $R$ is the mid-point of $DC$.
By Mid-point Theorem, $SR \parallel AC$ and $SR = \frac{1}{2} AC$.
$S$ is the mid-point of $DA$ and $R$ is the mid-point of $DC$.
By Mid-point Theorem, $SR \parallel AC$ and $SR = \frac{1}{2} AC$.
(ii) In $\triangle ABC$:
$P$ is the mid-point of $AB$ and $Q$ is the mid-point of $BC$.
By Mid-point Theorem, $PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
From (i), $SR = \frac{1}{2} AC$. Thus, $PQ = SR$.
$P$ is the mid-point of $AB$ and $Q$ is the mid-point of $BC$.
By Mid-point Theorem, $PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
From (i), $SR = \frac{1}{2} AC$. Thus, $PQ = SR$.
(iii) PQRS is a parallelogram:
From (i) and (ii), $PQ \parallel AC$ and $SR \parallel AC \Rightarrow PQ \parallel SR$.
Also $PQ = SR$.
A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.
From (i) and (ii), $PQ \parallel AC$ and $SR \parallel AC \Rightarrow PQ \parallel SR$.
Also $PQ = SR$.
A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.
Proved all parts.
Q2: Rhombus Mid-points
00:00
$ABCD$ 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.
From Q1, we know $PQRS$ is a parallelogram.
To prove it's a rectangle, we need to show one angle is $90^\circ$.
To prove it's a rectangle, we need to show one angle is $90^\circ$.
Diagonals of rhombus $ABCD$ intersect at $90^\circ$. Let diagonals intersect at $O$.
$AC \perp BD$.
$AC \perp BD$.
$PQ \parallel AC$ (Mid-point theorem).
$SP \parallel BD$ (Mid-point theorem).
Since $AC \perp BD$, then $PQ \perp SP$.
$\therefore \angle SPQ = 90^\circ$.
$SP \parallel BD$ (Mid-point theorem).
Since $AC \perp BD$, then $PQ \perp SP$.
$\therefore \angle SPQ = 90^\circ$.
A parallelogram with one right angle is a rectangle.
Proved.
Q3: Rectangle Mid-points
00:00
$ABCD$ 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.
$PQRS$ is a parallelogram (from Q1 logic).
To prove it's a rhombus, we need to show adjacent sides are equal ($PQ = QR$).
To prove it's a rhombus, we need to show adjacent sides are equal ($PQ = QR$).
Join diagonals $AC$ and $BD$. In a rectangle, $AC = BD$.
$PQ = \frac{1}{2} AC$ (Mid-point theorem in $\triangle ABC$).
$QR = \frac{1}{2} BD$ (Mid-point theorem in $\triangle BCD$).
Since $AC = BD$, then $PQ = QR$.
$QR = \frac{1}{2} BD$ (Mid-point theorem in $\triangle BCD$).
Since $AC = BD$, then $PQ = QR$.
A parallelogram with adjacent sides equal is a rhombus.
Proved.
Q5: Trisection of Diagonal
00:00
In a parallelogram $ABCD$, $E$ and $F$ are the mid-points of sides $AB$ and $CD$ respectively. Show that the line segments $AF$ and $EC$ trisect the diagonal $BD$.
Since $AB \parallel CD$ and $AB = CD$:
$AE = \frac{1}{2} AB$ and $CF = \frac{1}{2} CD$.
$\therefore AE = CF$ and $AE \parallel CF$.
Thus, $AECF$ is a parallelogram. So $AF \parallel EC$.
$AE = \frac{1}{2} AB$ and $CF = \frac{1}{2} CD$.
$\therefore AE = CF$ and $AE \parallel CF$.
Thus, $AECF$ is a parallelogram. So $AF \parallel EC$.
In $\triangle DQC$:
$F$ is mid-point of $DC$ and $FP \parallel CQ$ (part of $AF \parallel EC$).
By Converse of Mid-point Theorem, $P$ is mid-point of $DQ$.
$\therefore DP = PQ$.
$F$ is mid-point of $DC$ and $FP \parallel CQ$ (part of $AF \parallel EC$).
By Converse of Mid-point Theorem, $P$ is mid-point of $DQ$.
$\therefore DP = PQ$.
In $\triangle APB$:
$E$ is mid-point of $AB$ and $EQ \parallel AP$.
By Converse of Mid-point Theorem, $Q$ is mid-point of $PB$.
$\therefore PQ = QB$.
$E$ is mid-point of $AB$ and $EQ \parallel AP$.
By Converse of Mid-point Theorem, $Q$ is mid-point of $PB$.
$\therefore PQ = QB$.
From above, $DP = PQ = QB$.
Thus, $AF$ and $EC$ trisect $BD$.
Thus, $AF$ and $EC$ trisect $BD$.
Proved.
Q7: Right Triangle Mid-point
00:00
$ABC$ is a triangle right angled at $C$. A line through the mid-point $M$ of hypotenuse $AB$ and parallel to $BC$ intersects $AC$ at $D$. Show that:
(i) $D$ is the mid-point of $AC$
(ii) $MD \perp AC$
(iii) $CM = MA = \frac{1}{2} AB$
(i) $D$ is the mid-point of $AC$
(ii) $MD \perp AC$
(iii) $CM = MA = \frac{1}{2} AB$
(i) D is mid-point of AC:
In $\triangle ABC$, $M$ is mid-point of $AB$ and $MD \parallel BC$.
By Converse of Mid-point Theorem, $D$ is the mid-point of $AC$.
In $\triangle ABC$, $M$ is mid-point of $AB$ and $MD \parallel BC$.
By Converse of Mid-point Theorem, $D$ is the mid-point of $AC$.
(ii) $MD \perp AC$:
Since $MD \parallel BC$ and $AC$ is transversal, $\angle ADM = \angle ACB$ (Corresponding Angles).
Given $\angle ACB = 90^\circ$, so $\angle ADM = 90^\circ$.
Thus, $MD \perp AC$.
Since $MD \parallel BC$ and $AC$ is transversal, $\angle ADM = \angle ACB$ (Corresponding Angles).
Given $\angle ACB = 90^\circ$, so $\angle ADM = 90^\circ$.
Thus, $MD \perp AC$.
(iii) $CM = MA = \frac{1}{2} AB$:
In $\triangle ADM$ and $\triangle CDM$:
1. $AD = CD$ (D is mid-point)
2. $\angle ADM = \angle CDM = 90^\circ$
3. $DM = DM$ (Common)
$\therefore \triangle ADM \cong \triangle CDM$ (SAS Rule).
$\Rightarrow AM = CM$ (CPCT).
Since $M$ is mid-point of $AB$, $AM = \frac{1}{2} AB$.
Hence, $CM = MA = \frac{1}{2} AB$.
In $\triangle ADM$ and $\triangle CDM$:
1. $AD = CD$ (D is mid-point)
2. $\angle ADM = \angle CDM = 90^\circ$
3. $DM = DM$ (Common)
$\therefore \triangle ADM \cong \triangle CDM$ (SAS Rule).
$\Rightarrow AM = CM$ (CPCT).
Since $M$ is mid-point of $AB$, $AM = \frac{1}{2} AB$.
Hence, $CM = MA = \frac{1}{2} AB$.
Proved all parts.