NCERT Solutions Class 9 Maths Chapter 7 Triangles

5.0NCERT Questions with Solutions for Class 9 Maths Chapter 7 - Detailed Solutions

Exercise: 7.1

1. In quadrilateral $\mathrm{ACBD},\mathrm{AC}=\mathrm{AD}$ and AB bisects $\angle A$. Show that $△ABC\cong △ABD$. What can you say about BC and BD?

• Sol. Given : In quadrilateral $\mathrm{ACBD},\mathrm{AC}=\mathrm{AD}$ and AB bisect $\angle \mathrm{A}$. To prove : $△\mathrm{ABC}\cong △\mathrm{ABD}$ Proof : In $△ABC$ and $△ABD$ $\mathrm{AC}=\mathrm{AD}$ (Given) $\mathrm{AB}=\mathrm{AB}$ (Common) $\angle \mathrm{CAB}=\angle \mathrm{DAB}$ (AB bisect $\angle \mathrm{A}$ ) $\therefore △\mathrm{ABC}\cong △\mathrm{ABD}$ (by SAS criteria) $\mathrm{BC}=\mathrm{BD}$ (by CPCT)
• ABCD is a quadrilateral in which $\mathrm{AD}=\mathrm{BC}$ and $\angle \mathrm{DAB}=\angle \mathrm{CBA}$. Prove that

• (i) $△\mathrm{ABD}\cong △\mathrm{BAC}$ (ii) $\mathrm{BD}=\mathrm{AC}$ (iii) $\angle \mathrm{ABD}=\angle \mathrm{BAC}$ Sol. In $△ABD$ and $△BAC$, $\mathrm{AD}=\mathrm{BC}$ (Given) $\angle \mathrm{DAB}=\angle \mathrm{CBA}$ (Given) $\mathrm{AB}=\mathrm{AB}$ (Common side) $\therefore$ By SAS congruence rule, we have $△\mathrm{ABD}\cong △\mathrm{BAC}$ Also, by CPCT, we have $\mathrm{BD}=\mathrm{AC}$ and $\angle \mathrm{ABD}=\angle \mathrm{BAC}$
• $AD$ and $BC$ are equal perpendiculars to a line segment AB . Show that CD bisects AB .

• Sol. Given : AD and BC are equal perpendiculars to line $AB$. To prove : CD bisects AB Proof: In $△OAD$ and $△OBC$ $\mathrm{AD}=\mathrm{BC}$ (Given) $\angle \mathrm{OAD}=\angle \mathrm{OBC}($ Each $90^{\circ })$ $\angle \mathrm{AOD}=\angle \mathrm{BOC}$ (Vertically opposite angles) $\Delta \mathrm{OAD}\cong △\mathrm{OBC}$ (AAS rule) $OA=OB$ (by CPCT) $\therefore \mathrm{CD}$ bisects AB
• $\ell$ and $m$ are two parallel lines intersected by another pair of parallel lines p and q . Show that $△ABC△CDA$.

• Sol. In $△\mathrm{ABC}$ and $△\mathrm{CDA}$ $\angle \mathrm{CAB}=\angle \mathrm{ACD}$ (Pair of alternate angle) $\angle \mathrm{BCA}=\angle \mathrm{DAC}$ (Pair of alternate angle) $\mathrm{AC}=\mathrm{AC}$ (Common side) $\therefore △\mathrm{ABC}\cong △\mathrm{CDA}$ (ASA criteria)
• Line $\ell$ is the bisector of an angle $\angle \mathrm{A}$ and $B$ is any point on $\ell$. BP and BQ are perpendiculars from $B$ to the arms of $\angle A$. Show that:

• (i) $△\mathrm{APB}△\mathrm{AQB}$ (ii) $BP=BQ$ or $B$ is equidistant from the arms of $\angle \mathrm{A}$ Sol. Given : line $\ell$ is bisector of angle $A$ and $B$ is any point on $\ell$. BP and BQ are perpendicular from $B$ to arms of $\angle A$ To prove : (i) $\Delta \mathrm{APB}\cong △\mathrm{AQB}$ (ii) $\mathrm{BP}=\mathrm{BQ}$ Proof: (i) In $△\mathrm{APB}$ and $△\mathrm{AQB}$ $\angle \mathrm{BAP}=\angle \mathrm{BAQ}(\ell \text{ is bisector })$ $\mathrm{AB}=\mathrm{AB}\text{ (common) }$ $\angle \mathrm{BPA}=\angle \mathrm{BQA}($ Each $90^{\circ })$ $\therefore △\mathrm{APB}\cong △\mathrm{AQB}$ (AAS rule) (ii) $\Delta \mathrm{APB}\cong △\mathrm{AQB}$ $BP=BQ$ (By CPCT)
• In figure, $\mathrm{AC}=\mathrm{AE},\mathrm{AB}=\mathrm{AD}$ and $\angle \mathrm{BAD}=$ $\angle EAC$. Show that $BC=DE$.
  
  Sol. Given : $\mathrm{AC}=\mathrm{AE}$ $AB=AD$, $\angle \mathrm{BAD}=\angle \mathrm{EAC}$ To prove: $\mathrm{BC}=\mathrm{DE}$ Proof: In $△ABC$ and $△ADE$ $\mathrm{AB}=\mathrm{AD}$ (Given) $\mathrm{AC}=\mathrm{AE}$ (Given) $\angle \mathrm{BAD}=\angle \mathrm{EAC}$ Add $\angle \mathrm{DAC}$ to both $\Rightarrow \angle \mathrm{BAD}+\angle \mathrm{DAC}=\angle \mathrm{DAC}+\angle \mathrm{EAC}$ $\angle \mathrm{BAC}=\angle \mathrm{DAE}$ $△\mathrm{ABC}\cong △\mathrm{ADE}$ (SAS rule) $\mathrm{BC}=\mathrm{DE}$ (By CPCT)
• $AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle \mathrm{BAD}=\angle \mathrm{ABE}$ and $\angle \mathrm{EPA}=$ $\angle$ DPB. Show that (i) $△\mathrm{DAP}△\mathrm{EBP}$ (ii) $\mathrm{AD}=\mathrm{BE}$

• Sol. $\angle \mathrm{EPA}=\angle \mathrm{DPB}$ (Given) $\Rightarrow \angle \mathrm{EPA}+\angle \mathrm{DPE}=\angle \mathrm{DPB}+\angle \mathrm{DPE}$ $\Rightarrow \angle \mathrm{APD}=\angle \mathrm{BPE}$ Now, in $△$ DAP and $△EBP$, we have $\mathrm{AP}=\mathrm{PB}(\because \mathrm{P}$ is mid point of AB$)$ $\angle \mathrm{PAD}=\angle \mathrm{PBE}$ $\left\{\begin{array}{l}\because \angle \mathrm{PAD}=\angle \mathrm{BAD},\angle \mathrm{PBE}=\angle \mathrm{ABE}\\ \text{ and we are given that }\angle \mathrm{BAD}=\angle \mathrm{ABE}\end{array}\right\}$ Also, $\angle \mathrm{APD}=\angle \mathrm{BPE}$ $\therefore △\mathrm{DAP}\cong △\mathrm{EBP}($ By ASA congruence) $\Rightarrow \mathrm{AD}=\mathrm{BE}$ (By CPCT)
• In right triangle ABC , right angled at $\mathrm{C},\mathrm{M}$ is the mid-point of hypotenuse $AB.C$ is joined to M and produced to a point D such that $\mathrm{DM}=\mathrm{CM}$. Point D is joined to point B. Show that :

• (i) $\Delta \mathrm{AMC}\cong △\mathrm{BMD}$ (ii) $\angle \mathrm{DBC}$ is a right angle. (iii) $\Delta \mathrm{DBC}\cong △\mathrm{ACB}$ (iv) $\mathrm{CM}=\frac{1}{2}\mathrm{AB}$ Sol. (i) In $△AMC\cong △BMD$, $AM=BM(\because M$ is mid point of $AB)$ $\angle \mathrm{AMC}=\angle \mathrm{BMD}$ (Vertically opposite angles) $\mathrm{CM}=\mathrm{DM}$ (Given) $\therefore △\mathrm{AMC}\cong △\mathrm{BMD}$ (By SAS congruence) (ii) $\angle \mathrm{AMC}=\angle \mathrm{BMD}$, $\Rightarrow \angle \mathrm{ACM}=\angle \mathrm{BDM}$ (By CPCT) $\Rightarrow \mathrm{CA}\Vert \mathrm{BD}$ $\Rightarrow \angle \mathrm{BCA}+\angle \mathrm{DBC}=180^{\circ }$ $\Rightarrow \angle \mathrm{DBC}=90^{\circ }$ $\left(\because \angle \mathrm{BCA}=90^{\circ }\right)$ (iii) In $△\mathrm{DBC}$ and $△\mathrm{ACB}$, $\mathrm{DB}=\mathrm{AC}(\because \Delta \mathrm{BMD}\cong △\mathrm{AMC})$ $\begin{array}{ccc}& \angle \mathrm{DBC}=\angle \mathrm{ACB}& \left(\text{ Each }=90^{\circ }\right)\\ & \mathrm{BC}=\mathrm{CB}& (\text{ Common side) }\\ \therefore & △\mathrm{DBC}\cong △\mathrm{ACB}& \text{ (By SAS congruence) }\end{array}$ (iv) $\mathrm{In}△\mathrm{DBC}\cong △\mathrm{ACB}$ $\Rightarrow \mathrm{CD}=\mathrm{AB}$ $\Rightarrow \mathrm{CM}=\mathrm{DM}$ (given) $\Rightarrow \mathrm{CM}=\mathrm{DM}=\frac{1}{2}\mathrm{CD}$ $\Rightarrow \mathrm{CD}=2\mathrm{CM}$ From (1) and (2), $2\mathrm{CM}=\mathrm{AB}$ $\Rightarrow \mathrm{CM}=\frac{1}{2}\mathrm{AB}$

Exercise : 7.2

• In an isosceles triangle ABC , with $\mathrm{AB}=\mathrm{AC}$, the bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$ intersect each other at O . Join A to O . Show that: (i) $\mathrm{OB}=\mathrm{OC}$ (ii) AO bisects $\angle \mathrm{A}$. Sol.

• (i) In $△\mathrm{ABC},\mathrm{OB}$ and OC are bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$. $\therefore \angle \mathrm{OBC}=\frac{1}{2}\angle \mathrm{B}$ $\angle \mathrm{OCB}=\frac{1}{2}\angle \mathrm{C}$ Also, $\mathrm{AB}=\mathrm{AC}$ (Given) $\Rightarrow \angle \mathrm{B}=\angle \mathrm{C}$ From (1), (2), (3), we have $\angle \mathrm{OBC}=\angle \mathrm{OCB}$ Now, in $△\mathrm{OBC}$, we have $\angle \mathrm{OBC}=\angle \mathrm{OCB}$ $\Rightarrow \mathrm{OB}=\mathrm{OC}$ (Sides opposite to equal angles are equal) (ii) $\angle \mathrm{OBA}=\frac{1}{2}\angle \mathrm{B}$ and $\angle \mathrm{OCA}=\frac{1}{2}\angle \mathrm{C}$ $\Rightarrow \angle \mathrm{OBA}=\angle \mathrm{OCA}(\because \angle \mathrm{B}=\angle \mathrm{C})$ $\mathrm{AB}=\mathrm{AC}$ and $\mathrm{OB}=\mathrm{OC}$ $\therefore △\mathrm{OAB}\cong △\mathrm{OAC}$ (SAS congruence criteria) $\Rightarrow \angle \mathrm{OAB}=\angle \mathrm{OAC}$ $\Rightarrow$ AO bisects $\angle A$.
• In $△\mathrm{ABC},\mathrm{AD}$ is the perpendicular bisector of $BC$. Show that $△ABC$ is an isosceles triangle in which $AB=AC$.

• Sol. Given : In $△\mathrm{ABC},\mathrm{AD}$ is perpendicular bisector of BC. To Prove : $△\mathrm{ABC}$ is isosceles $\Delta$ with Proof: In $△ADB$ and $△ADC$ $\angle \mathrm{ADB}=\angle \mathrm{ADC}$ (Each $90^{\circ }$ ) $\mathrm{DB}=\mathrm{DC}(\mathrm{AD}$ is $\perp$ bisector of BC$)$ $\mathrm{AD}=\mathrm{AD}$ (Common) $△\mathrm{ADB}\cong △\mathrm{ADC}$ (By SAS rule) $\mathrm{AB}=\mathrm{AC}$ (By CPCT) $\therefore △\mathrm{ABC}$ is an isosceles $\Delta$ with $\mathrm{AB}=\mathrm{AC}$
• ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides $AC$ and $AB$ respectively. Show that these altitudes are equal.

• Sol. In $△ABE$ and $△ACF$, we have $\angle \mathrm{BEA}=\angle \mathrm{CFA}$ $($ Each $=90^{\circ })$ $\angle \mathrm{A}=\angle \mathrm{A}$ (Common angle) $\mathrm{AB}=\mathrm{AC}$ (Given) $\therefore △\mathrm{ABE}\cong △\mathrm{ACF}$ (By AAS congruence criteria) $\Rightarrow \mathrm{BE}=\mathrm{CF}$ (By CPCT)
• ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (See figure). Show that (i) $△\mathrm{ABE}\cong △\mathrm{ACF}$ (ii) $\mathrm{AB}=\mathrm{AC}$, i.e., ABC is an isosceles triangle.
  
  Sol. (i) In $△ABE$ and $△ACF$, we have $\angle \mathrm{A}=\angle \mathrm{A}$ (Common) $\angle \mathrm{AEB}=\angle \mathrm{AFC}($ Each $=90^{\circ })$ $\mathrm{BE}=\mathrm{CF}$ (Given) $\therefore △\mathrm{ABE}\cong △\mathrm{ACF}$ (By AAS congruence) (ii) $△\mathrm{ABE}\cong △\mathrm{ACF}$ $\Rightarrow \mathrm{AB}=\mathrm{AC}$ (By CPCT)
• ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that $\angle \mathrm{ABD}=\angle \mathrm{ACD}$.

• Sol. Given : ABC and BCD are two isosceles triangle on common base BC. To prove: $\angle \mathrm{ABC}=\angle \mathrm{ACD}$ Proof: ABC is an isosceles Triangle on base BC $\therefore \angle \mathrm{ABC}=\angle \mathrm{ACB}$ $\because DBC$ is an isosceles $\Delta$ on base $BC$. $\angle \mathrm{DBC}=\angle \mathrm{DCB}$ Adding (1) and (2) $\angle \mathrm{ABC}+\angle \mathrm{DBC}=\angle \mathrm{ACB}+\angle \mathrm{DCB}$ $\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{ACD}$
• $△\mathrm{ABC}$ is an isosceles triangle in which $AB=AC$. Side BA is produced to $D$ such that $AD=AB$ (see figure). Show that $\angle \mathrm{BCD}$ is a right angle.

• Sol. In $△ABC,AB=AC$ $\Rightarrow \angle \mathrm{ACB}=\angle \mathrm{ABC}$ In $△\mathrm{ACD}$, $\mathrm{AD}=\mathrm{AB}$ (By construction) $\Rightarrow \mathrm{AD}=\mathrm{AC}$ $\Rightarrow \angle \mathrm{ACD}=\angle \mathrm{ADC}$ Adding (1) and (2), $\angle \mathrm{ACB}+\angle \mathrm{ACD}=\angle \mathrm{ABC}+\angle \mathrm{ADC}$ $\Rightarrow \angle \mathrm{BCD}=\angle \mathrm{ABC}+\angle \mathrm{ADC}$ In $△\mathrm{DBC}$, $\angle \mathrm{ABC}+\angle \mathrm{BCD}+\angle \mathrm{CDB}=180^{\circ }$ $\Rightarrow 2\angle \mathrm{BCD}=180^{\circ }$ $\Rightarrow \angle \mathrm{BCD}=90^{\circ }$
• ABC is a right angled triangle in which $\angle \mathrm{A}$ $=90^{\circ }$ and $AB=AC$. Find $\angle B$ and $\angle C$. Sol.

• In $△\mathrm{ABC}$ $\mathrm{AB}=\mathrm{AC}$ $\angle \mathrm{B}=\angle \mathrm{C}$ (angles opposite to equal sides are equal) In $△\mathrm{ABC}$ $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }$ $90^{\circ }+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }$ $\angle \mathrm{B}+\angle \mathrm{C}=90^{\circ }$ from (1) and (2) $\angle \mathrm{B}=\angle \mathrm{C}=45^{\circ }$
• Show that the angles of an equilateral triangle are $60^{\circ }$ each. Sol.

• $△\mathrm{ABC}$ is equilateral triangle. $\Rightarrow \mathrm{AB}=\mathrm{BC}=\mathrm{CA}$ Now, $\mathrm{AB}=\mathrm{BC}$ $\Rightarrow \mathrm{BA}=\mathrm{BC}$ $\Rightarrow \angle \mathrm{C}=\angle \mathrm{A}$ Similarly, $\angle \mathrm{A}=\angle \mathrm{B}$ From (1) and (2), $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}$ Also, $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }\dots$ (4) $\Rightarrow \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\frac{1}{3}\times 180^{\circ }=60^{\circ }$

Exercise: 7.3

• $△\mathrm{ABC}$ and $△\mathrm{DBC}$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of BC (see figure). If AD is extended to intersect BC at P , show that

• (i) $△\mathrm{ABD}\cong △\mathrm{ACD}$ (ii) $△\mathrm{ABP}\cong △\mathrm{ACP}$ (iii) AP bisects $\angle \mathrm{A}$ as well as $\angle \mathrm{D}$ (iv) AP is the perpendicular bisector of BC . Sol. (i) In $△ABD$ and $△ACD$, $$\begin{gathered} \mathrm{AB}=\mathrm{AC}(\because \Delta \mathrm{ABC}\text{ is isosceles }) \\ \mathrm{DB}=\mathrm{DC}(\because \Delta \mathrm{DBC}\text{ is isosceles }) \end{gathered}$$ (iv) $△\mathrm{ABP}\cong △\mathrm{ACP}$ (Proved above) $\Rightarrow \mathrm{BP}=\mathrm{CP}$ (By CPCT) $\Rightarrow \mathrm{AP}$ bisects BC $\angle \mathrm{APB}=\angle \mathrm{APC}$ (By CPCT) $\angle \mathrm{APB}+\angle \mathrm{APC}=180^{\circ }$ $2\angle \mathrm{APB}=180^{\circ }$ $\angle \mathrm{APB}=90^{\circ }$ AP is perpendicular bisector of $BC$
• $AD$ is an altitude of an isosceles triangle $ABC$ in which $AB=AC$. Show that (i) AD bisects BC (ii) AD bisects $\angle \mathrm{A}$ Sol. Given : AD is an altitude of an isosceles triangle $ABC$ in which $AB=AC$.

• To Prove: (i) AD bisect BC . (ii) AD bisect $\angle \mathrm{A}$. Proof: (i) In right $△ADB$ and right $△ADC$. Hyp. $AB=$ Hyp. $AC$ $\angle \mathrm{ADB}=\angle \mathrm{ADC}($ Each $90^{\circ })$ Side AD = side AD (Common) $\Delta \mathrm{ADB}\cong △\mathrm{ADC}$ $\Rightarrow \mathrm{BD}=\mathrm{CD}$ $\Rightarrow \mathrm{AD}$ bisect BC (ii) $△\mathrm{ADB}\cong △\mathrm{ADC}$ $\angle \mathrm{BAD}=\angle \mathrm{CAD}$ (By CPCT) $\Rightarrow \mathrm{AD}$ bisect $\angle \mathrm{A}$
• Two sides AB and BC and median AM of one triangle $ABC$ are respectively equal to sides PQ and QR and median PN of $△\mathrm{PQR}$ (see figure). Show that :

• (i) $△\mathrm{ABM}\cong △\mathrm{PQN}$ (ii) $△\mathrm{ABC}\cong △\mathrm{PQR}$ Sol. (i) $\mathrm{BM}=\frac{1}{2}\mathrm{BC}(\because \mathrm{M}$ is mid-point of BC$)$ $\mathrm{QN}=\frac{1}{2}\mathrm{QR}(\because \mathrm{N}$ is mid-point of QR$)$ $\Rightarrow BM=QN(\because BC=QR$ is given $)$ Now, in $△ABM$ and $△PQN$, we have $AB=PQ$ (Given) $BM=QN$ (Proved) $\mathrm{AM}=\mathrm{PN}$ (Given) $\therefore △\mathrm{ABM}\cong △\mathrm{PQN}$ (SSS congruence criteria) (ii) $△\mathrm{ABM}\cong △\mathrm{PQN}$ $\Rightarrow \angle \mathrm{ABM}=\angle \mathrm{PQN}$ $\angle \mathrm{B}=\angle \mathrm{Q}$ (By CPCT) Now, in $△\mathrm{ABC}$ and $△\mathrm{PQN}$, $\mathrm{AB}=\mathrm{PQ},\angle \mathrm{B}=\angle \mathrm{Q}$ and $\mathrm{BC}=\mathrm{QR}$ $\Rightarrow △\mathrm{ABC}\cong △\mathrm{PQN}$ [by SAS congruence]
• BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle $ABC$ is isosceles. Sol. Given BE and CF are two altitude of $△\mathrm{ABC}$.

• To prove : $△\mathrm{ABC}$ is isosceles. Proof : In right $△\mathrm{BEC}$ and right $△\mathrm{CFB}$ side $\mathrm{BE}=$ side $\mathrm{CF}$ (Given) Hyp.BC = Hyp CB (Common) $\angle \mathrm{BEC}=\angle \mathrm{BFC}($ Each $90^{\circ })$ $△\mathrm{BEC}\cong △\mathrm{CFB}$ (RHS Rule) $\therefore \angle \mathrm{BCE}=\angle \mathrm{CBF}$ (By CPCT) $\mathrm{AB}=\mathrm{AC}$ (Side opp. to equal angles are equal) $△\mathrm{ABC}$ is isosceles.
• ABC is an isosceles triangle with $\mathrm{AB}=\mathrm{AC}$. Draw $\mathrm{AP}\perp \mathrm{BC}$ to show that $\angle \mathrm{B}=\angle \mathrm{C}$ Sol.

• In $△APB$ and $△APC$ $AB=AC$ (Given) $\angle \mathrm{APB}=\angle \mathrm{APC}$ (Each $=90^{\circ }$ ) $\mathrm{AP}=\mathrm{AP}$ (common side) Therefore, by RHS congruence criteria, we have $\Delta APB\cong △APC$ $\Rightarrow \angle \mathrm{ABP}=\angle \mathrm{ACP}$ (By CPCT) $\Rightarrow \angle B=\angle C$

Exercise : 7.4

• Show that in a right angled triangle, the hypotenuse is the longest side. Sol.

• $△ABC$ is right angled at $B$. AC is hypotenuse. Now, $\angle B=90^{\circ }$ and $\angle \mathrm{A}+\angle \mathrm{C}=90^{\circ }$ $\Rightarrow \angle \mathrm{A}<90^{\circ }$ and $\angle \mathrm{C}<90^{\circ }$ $\Rightarrow \angle \mathrm{B}>\angle \mathrm{A}$ and $\angle \mathrm{B}>\angle \mathrm{C}$ $\Rightarrow \mathrm{AC}>\mathrm{BC}$ and $\mathrm{AC}>\mathrm{AB}$. $\therefore$ Hypotenuse AC is the longest side of the right angled $△\mathrm{ABC}$.
• In figure, sides $AB$ and $AC$ of $△ABC$ are extended to points P and Q respectively. Also, $\angle \mathrm{PBC}<\angle \mathrm{QCB}$. Show that $\mathrm{AC}>\mathrm{AB}$

• Sol. Sides $AB$ and $AC$ of $△ABC$ are extended to points $P$ and $Q$ To prove : AC > AB Proof: $\angle \mathrm{PBC}<\angle \mathrm{QCB}$ (Given) $180^{\circ }-\angle \mathrm{PBC}>180^{\circ }-\angle \mathrm{QCB}$ $\angle \mathrm{ABC}>\angle \mathrm{ACB}$ $\Rightarrow \mathrm{AC}>\mathrm{AB}$ (sides opposite to greater angle is longer)
• In fig, $\angle \mathrm{B}<\angle \mathrm{A}$ and $\angle \mathrm{C}<\angle \mathrm{D}$. Show that $\mathrm{AD}<\mathrm{BC}$.

• Sol. $\angle B<\angle A$ in $△OAB$ $\Rightarrow OA<OB$ Also, $\angle \mathrm{C}<\angle \mathrm{D}$ in $△\mathrm{OCD}$ $\Rightarrow \mathrm{OD}<\mathrm{OC}$ Adding (1) and (2), $OA+OD<OB+OC$ $\Rightarrow \mathrm{AD}<\mathrm{BC}$
• AB and CD are respectively the smallest and longest sides of a quadrilateral $ABCD$ (see figure). Show that $\angle \mathrm{A}>\angle \mathrm{C}$ and $\angle B>\angle D$.

• Sol.

• In quadrilateral $\mathrm{ABCD},\mathrm{AB}$ is the smallest side and $CD$ is the longest side. Join $AC$ and BD . In $△\mathrm{ABC}$, $\mathrm{BC}>\mathrm{AB}$ $\Rightarrow \angle 1>\angle 3$ In $△ACD$, $\mathrm{CD}>\mathrm{AD}(\because \mathrm{CD}$ is longest side $)$ $\Rightarrow \angle 2>\angle 4$ Adding (1) and (2), we have $\angle 1+\angle 2>\angle 3+\angle 4$ $\Rightarrow \angle \mathrm{A}>\angle \mathrm{C}$ Similarly, we can prove that $\angle \mathrm{B}>\angle \mathrm{D}$
• In figure, $\mathrm{PR}>\mathrm{PQ}$ and PS bisects $\angle \mathrm{QPR}$. Prove that $\angle \mathrm{PSR}>\angle \mathrm{PSQ}$.

• Sol. Given : PR > PQ, PS bisect $\angle \mathrm{QPR}$. To prove : $\angle \mathrm{PSR}>\angle \mathrm{PSQ}$ Proof: In $△\mathrm{PQR}$ PR > PQ (Given) $\angle \mathrm{PQR}>\angle \mathrm{PRQ}$ (angle opposite to longer side is greater) and $\angle \mathrm{PQS}>\angle \mathrm{PRS}$ PS bisects $\angle \mathrm{QPR}$ $\Rightarrow \angle \mathrm{QPS}=\angle \mathrm{RPS}$ From (1) and (2) $\angle \mathrm{PQS}+\angle \mathrm{QPS}>\angle \mathrm{PRS}+\angle \mathrm{RPS}$ $\angle \mathrm{PSR}>\angle \mathrm{PSQ}$ (by exterior angle property) Hence proved.
• Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest. Sol. Let us have AB as the perpendicular line segment and AP is any other line segment. Now $△ABP$ is right angled and AP is hypotenuse. Here, $\angle \mathrm{B}>\angle \mathrm{P}\left(\because \angle \mathrm{B}=90^{\circ }\right)$ $\Rightarrow \mathrm{AP}>\mathrm{AB}$

• Thus, perpendicular line segment is smallest.