NCERT Solutions for Class 9 Maths Chapter 12 — Exercise Set 12.3#
Theorem 6 (Midpoint Theorem): the segment joining the midpoints of two sides of a triangle is parallel to the third side and has half its length.
Theorem 7 (Converse): the line through the midpoint of one side, parallel to another side, bisects the third side.
Theorem 9 (Varignon): the midpoints of the four sides of a quadrilateral are the vertices of a parallelogram.
1. (i) If P, Q, R are the midpoints of sides AB, AC, BC respectively of ΔABC, show that ΔPQR is congruent to ΔQPA and to two other triangles which you should identify. (ii) Suppose someone erases ΔABC, leaving only ΔPQR on the paper. Can you reconstruct ΔABC from ΔPQR? #
(i) Four congruent triangles
By the Midpoint Theorem in $\triangle ABC$:
$$PQ = \dfrac{1}{2}BC, \quad QR = \dfrac{1}{2}AB, \quad RP = \dfrac{1}{2}AC$$Since $P$, $Q$, $R$ are midpoints, the half-sides are:
$$BR = RC = \dfrac{1}{2}BC, \quad AP = PB = \dfrac{1}{2}AB, \quad AQ = QC = \dfrac{1}{2}AC$$So each small triangle has sides $\dfrac{1}{2}BC$, $\dfrac{1}{2}AB$, $\dfrac{1}{2}AC$:
| Triangle | Sides |
|---|---|
| $\triangle PQR$ | $PQ = \dfrac{BC}{2}$, $QR = \dfrac{AB}{2}$, $RP = \dfrac{AC}{2}$ |
| $\triangle QPA$ | $QP = \dfrac{BC}{2}$, $PA = \dfrac{AB}{2}$, $AQ = \dfrac{AC}{2}$ |
| $\triangle RBP$ | $RB = \dfrac{BC}{2}$, $BP = \dfrac{AB}{2}$, $PR = \dfrac{AC}{2}$ |
| $\triangle CRQ$ | $CR = \dfrac{BC}{2}$, $RQ = \dfrac{AB}{2}$, $QC = \dfrac{AC}{2}$ |
By SSS, $\triangle PQR \cong \triangle QPA \cong \triangle RBP \cong \triangle CRQ$. (Each is also a half-size copy of $\triangle ABC$.)
(ii) Reconstructing ΔABC
Yes. By the Midpoint Theorem, each side of $\triangle ABC$ passes through a vertex of $\triangle PQR$ and is parallel to the opposite side of $\triangle PQR$: $AB \parallel QR$ through $P$, $AC \parallel PR$ through $Q$, $BC \parallel PQ$ through $R$.
So: through each vertex of $\triangle PQR$, draw the line parallel to the opposite side. The three lines meet in pairs at $A$, $B$ and $C$.
2. In ΔABC, let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD. #
Proof
Let $MN$ meet $AD$ at $X$. By the Midpoint Theorem in $\triangle ABC$, $MN \parallel BC$, so $MX \parallel BD$.
In $\triangle ABD$, $M$ is the midpoint of $AB$, and the line $MX$ through $M$ is parallel to $BD$. By Theorem 7, it bisects the third side $AD$: $X$ is the midpoint of $AD$.
∴ $MN$ bisects $AD$.
3. In a quadrilateral ABCD, suppose AB ∥ DC and AB ≠ CD. Suppose G and H are the midpoints of AC and BD respectively. Prove that GH ∥ AB. (Why did we assume AB ≠ CD?) #
Proof
Let $E$ be the midpoint of $AD$.
In $\triangle ADC$, $E$ and $G$ are the midpoints of $AD$ and $AC$, so by the Midpoint Theorem:
$$EG \parallel DC$$In $\triangle DAB$, $E$ and $H$ are the midpoints of $DA$ and $DB$, so:
$$EH \parallel AB$$But $AB \parallel DC$, so $EG$ and $EH$ are both lines through $E$ parallel to $AB$. There is only one such line, so $E$, $G$, $H$ lie on it.
∴ $GH$ lies on a line parallel to $AB$, that is, $GH \parallel AB$.
Why AB ≠ CD: if $AB = CD$ as well as $AB \parallel CD$, then $ABCD$ is a parallelogram (Theorem 5). Its diagonals bisect each other, so the midpoints $G$ and $H$ are the same point, and there is no segment $GH$.
4. Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively. #
(i) Show that PR and QS bisect each other.
By Theorem 9, $PQRS$ is a parallelogram. $PR$ and $QS$ are its diagonals, and the diagonals of a parallelogram bisect each other (Theorem 1(c)).
(ii) Show that if AC = BD, then PR and QS are perpendicular. Is the converse true?
By the Midpoint Theorem:
$$PQ = \dfrac{1}{2}AC \quad (\text{in } \triangle ABC)$$$$QR = \dfrac{1}{2}BD \quad (\text{in } \triangle BCD)$$
If $AC = BD$, then $PQ = QR$. A parallelogram with two adjacent sides equal is a rhombus, and the diagonals of a rhombus are perpendicular. So $PR \perp QS$.
Converse — true. If $PR \perp QS$, the parallelogram $PQRS$ has perpendicular diagonals, so it is a rhombus (Exercise Set 12.2, Q3(ii)). Then $PQ = QR$, which gives $\dfrac{1}{2}AC = \dfrac{1}{2}BD$, so $AC = BD$.
5. Suppose PQRS is the Varignon parallelogram of ABCD. (i) Copy only PQRS on another paper. Show how you will recreate a congruent copy A′B′C′D′ of ABCD from PQRS. (ii) There are multiple ways to construct A′B′C′D′ in (i). Justify why the quadrilateral A′B′C′D′ you constructed is congruent to ABCD. (iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove the converse. #
(i) Construction
$P$, $Q$, $R$, $S$ are the midpoints of $AB$, $BC$, $CD$, $DA$. Each vertex of $ABCD$ is “reflected through a midpoint” to get the next vertex.
- Place A′: copy the position of $A$ relative to the parallelogram — for example, copy the triangle $SAP$ (measure $SA$ and $PA$ on the original and draw arcs from $S$ and $P$; take the point on the same side as $A$).
- Place B′: extend $A'P$ beyond $P$ by its own length: $PB' = A'P$. Then $P$ is the midpoint of $A'B'$.
- Place C′: extend $B'Q$ beyond $Q$ so that $QC' = B'Q$.
- Place D′: extend $C'R$ beyond $R$ so that $RD' = C'R$.
Join $A'B'C'D'$.
(ii) Why A′B′C′D′ is congruent to ABCD
Step 1. $\triangle SA'P \cong \triangle SAP$ (copied, SSS), so $A'$ is placed exactly where $A$ was relative to $PQRS$.
Step 2. $B$ is the unique point with $P$ as the midpoint of $AB$, and $B'$ is the unique point with $P$ as the midpoint of $A'B'$. Since $A'$ is in the position of $A$, $B'$ is in the position of $B$. In the same way, $C'$ is in the position of $C$, and $D'$ in the position of $D$.
Step 3 (closing up). We must also check that the last side $D'A'$ has $S$ as its midpoint. $D'$ and $A'$ are in the positions of $D$ and $A$, and $S$ is the midpoint of $DA$; equivalently, $\triangle SDR \cong \triangle SD'R$ (SSS, as $D'$ occupies the position of $D$), so $SD' = SD = SA = SA'$ and $S$ lies on $D'A'$.
So every vertex and side of $A'B'C'D'$ matches $ABCD$, and the two are congruent.
(iii) PQRS is a square ⟺ AC ⊥ BD and AC = BD
By the Midpoint Theorem:
$$PQ \parallel AC, \quad PQ = \dfrac{1}{2}AC$$$$QR \parallel BD, \quad QR = \dfrac{1}{2}BD$$
If PQRS is a square: $PQ = QR$ gives $AC = BD$; and $PQ \perp QR$ gives $AC \perp BD$ (lines parallel to perpendicular lines are perpendicular).
Converse — true: if $AC = BD$ and $AC \perp BD$, then $PQ = QR$ and $PQ \perp QR$. A parallelogram with two equal adjacent sides meeting at a right angle is a square.