NCERT Solutions · Class 9 Mathematics Quadrilaterals

(i) In the given figure, the Midpoint Theorem gives \[PQ = \tfrac12 BC,\quad QR = \tfrac12 AB,\quad RP = \tfrac12 AC \] and P, Q, R are midpoints, so \[AP = PB = \tfrac12 AB,\quad AQ = QC = \tfrac12 AC,\quad BR = RC = \tfrac12 BC \] By SSS: \[PQ = QP,\ QR = PA,\ RP = AQ \;\Rightarrow\; \triangle PQR \cong \triangle QPA \] \[PQ = RB,\ QR = BP,\ RP = PR \;\Rightarrow\; \triangle PQR \cong \triangle RBP \] \[PQ = CR,\ QR = RQ,\ RP = QC \;\Rightarrow\; \triangle PQR \cong \triangle CRQ \](ii) Yes. Through P, Q, R draw lines \(\displaystyle \ell_P \parallel QR\), \(\displaystyle \ell_Q \parallel RP\), \(\displaystyle \ell_R \parallel PQ\), and let \[A = \ell_P \cap \ell_Q,\quad B = \ell_P \cap \ell_R,\quad C = \ell_Q \cap \ell_R \] APRQ and PBRQ are parallelograms, so \[AP = QR = PB \] with A, P, B on \(\displaystyle \ell_P\): P is the midpoint of AB. Likewise Q is the midpoint of AC and R of BC.Answer: (i) \(\displaystyle \triangle PQR \cong \triangle QPA \cong \triangle RBP \cong \triangle CRQ\). (ii) Yes: the lines through P, Q, R parallel to QR, RP, PQ bound \(\displaystyle \triangle ABC\).

Let the line MN meet AD at E. \[MN \parallel BC \quad \text{(Midpoint Theorem, } \triangle ABC) \] \[ME \parallel BD,\ AM = MB \;\Rightarrow\; AE = ED,\ ME = \tfrac12 BD \quad \text{(Theorem 7, } \triangle ABD) \] \[NE \parallel DC,\ AN = NC \;\Rightarrow\; EN = \tfrac12 DC \quad \text{(Theorem 7, } \triangle ACD) \] \[ME + EN = \tfrac12 (BD + DC) = \tfrac12 BC = MN \] So E lies on segment MN and is the midpoint of AD.Answer: MN passes through the midpoint E of AD, so MN bisects AD.

Extend AH to meet line DC at K.\[BH = DH \quad \text{(H is the midpoint of BD)} \] \[\angle ABH = \angle KDH \quad \text{(alternate angles, } AB \parallel DC) \] \[\angle AHB = \angle KHD \quad \text{(vertically opposite angles)} \] \[\triangle ABH \cong \triangle KDH \ \text{(ASA)} \;\Rightarrow\; AH = HK,\quad DK = AB \] In \(\displaystyle \triangle ACK\), G and H are the midpoints of AC and AK, so \[GH \parallel CK \ \text{(Midpoint Theorem)}, \ \text{i.e. } GH \parallel DC \parallel AB \] \[GH = \tfrac12 CK = \tfrac12 |CD - AB| \] If AB = CD, then ABCD is a parallelogram (Theorem $\displaystyle 5$) and its diagonals bisect each other, so G = H and there is no line GH.Answer: \(\displaystyle GH \parallel AB\) and \(\displaystyle GH = \tfrac12 |CD - AB|\); AB \(\displaystyle \neq\) CD is assumed so that G and H are different points.

(i) Midpoint Theorem in \(\displaystyle \triangle ABC\) and \(\displaystyle \triangle ADC\): \[PQ \parallel AC \parallel SR,\qquad PQ = \tfrac12 AC = SR \] So PQRS is a parallelogram (Theorem $\displaystyle 5$). Its diagonals bisect each other (Theorem $\displaystyle 1$); with O their crossing point: \[OP = OR,\quad OQ = OS \](ii) Midpoint Theorem in \(\displaystyle \triangle BCD\) and \(\displaystyle \triangle ABD\): \(\displaystyle QR = PS = \tfrac12 BD\). If \(\displaystyle AC = BD\): \[PQ = \tfrac12 AC = \tfrac12 BD = QR \] \[PQ = RQ,\ OP = OR,\ OQ = OQ \;\Rightarrow\; \triangle POQ \cong \triangle ROQ \ \text{(SSS)} \] \[\angle POQ = \angle ROQ,\quad \angle POQ + \angle ROQ = 180^\circ \;\Rightarrow\; \angle POQ = 90^\circ \] Converse: True. If \(\displaystyle PR \perp QS\): \[OQ = OS,\ \angle POQ = \angle POS = 90^\circ,\ OP = OP \;\Rightarrow\; \triangle POQ \cong \triangle POS \ \text{(SAS)} \] \[PQ = PS \;\Rightarrow\; \tfrac12 AC = \tfrac12 BD \;\Rightarrow\; AC = BD \]Answer: (i) PR and QS bisect each other. (ii) \(\displaystyle AC = BD \Rightarrow PR \perp QS\); the converse is true.

(i) Copy PQRS. Trace the four corner triangles of ABCD and place each on its side of PQRS, outside it: \[\triangle A'SP \cong \triangle ASP,\ \ \triangle B'PQ \cong \triangle BPQ,\ \ \triangle C'QR \cong \triangle CQR,\ \ \triangle D'RS \cong \triangle DRS \] Join A', B', C', D' (right-hand figure).(ii) A, S, D are collinear, and the corner triangles are congruent, so \[\angle A'SP + \angle PSR + \angle RSD' = \angle ASP + \angle PSR + \angle RSD = 180^\circ \] \[SA' = SA = SD = SD' \] So S is the midpoint of A'D'; the same holds at P, Q, R. Then \[A'B' = A'P + PB' = AP + PB = AB,\quad B'C' = BC,\ C'D' = CD,\ D'A' = DA \] \[\angle D'A'B' = \angle SA'P = \angle SAP = \angle DAB,\ \text{and so at } B', C', D' \] Sides and angles match, so A'B'C'D' \(\displaystyle \cong\) ABCD.(iii) Midpoint Theorem: \[PQ \parallel AC,\ PQ = \tfrac12 AC;\qquad QR \parallel BD,\ QR = \tfrac12 BD \]If PQRS is a square: \[QR \perp PQ,\ PQ \parallel AC \;\Rightarrow\; QR \perp AC;\qquad BD \parallel QR \;\Rightarrow\; BD \perp AC \] \[\tfrac12 AC = PQ = QR = \tfrac12 BD \;\Rightarrow\; AC = BD \] Converse: let \(\displaystyle AC = BD\) and \(\displaystyle AC \perp BD\). \[PQ = SR = \tfrac12 AC = \tfrac12 BD = QR = PS \] \[PQ \parallel AC,\ QR \parallel BD,\ AC \perp BD \;\Rightarrow\; \angle PQR = 90^\circ \] A rhombus with a right angle is a square.Answer: (i) Fit the four corner triangles of ABCD onto PQRS. (ii) Collinear at S, P, Q, R with equal sides and angles, so A'B'C'D' \(\displaystyle \cong\) ABCD. (iii) PQRS is a square exactly when \(\displaystyle AC = BD\) and \(\displaystyle AC \perp BD\).