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NCERT Exemplar · Class 9 Mathematics Triangles

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EXERCISE 7.3 1–11 (part 3 of 5)

  1. Exercise 1

    ABC is an isosceles triangle with AB=AC\displaystyle \mathrm{AB}=\mathrm{AC} and BD and CE are its two medians. Show that BD=CE\displaystyle \mathrm{BD}=\mathrm{CE}.

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    \[AB=AC \quad \text{(given)} \] \[AD=\tfrac12 AC,\ AE=\tfrac12 AB \quad \text{(D, E are midpoints)} \] \[\Rightarrow AD=AE \] In \(\displaystyle \Delta ABD\) and \(\displaystyle \Delta ACE\): \[AB=AC,\ \angle A=\angle A,\ AD=AE \] \[\Rightarrow \Delta ABD \cong \Delta ACE \quad \text{(SAS)} \] \[\Rightarrow BD=CE \quad \text{(CPCT)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q1 Answer: BD = CE.

  2. Exercise 2

    In Fig.7.4, D and E are points on side BC of a ΔABC\displaystyle \Delta \mathrm{ABC} such that BD=CE\displaystyle \mathrm{BD}=\mathrm{CE} and AD=AE\displaystyle \mathrm{AD}=\mathrm{AE}. Show that ΔABD≅△ACE\displaystyle \Delta \mathrm{ABD} \cong \triangle \mathrm{ACE}. NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-3_Q2

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    \[AD=AE \Rightarrow \angle ADE=\angle AED \quad \text{(isosceles } \Delta ADE\text{)} \] \[\angle ADB=180^\circ-\angle ADE,\ \angle AEC=180^\circ-\angle AED \quad \text{(linear pairs on BC)} \] \[\Rightarrow \angle ADB=\angle AEC \] In \(\displaystyle \Delta ABD\) and \(\displaystyle \Delta ACE\): \[BD=CE,\ \angle ADB=\angle AEC,\ AD=AE \] \[\Rightarrow \Delta ABD \cong \Delta ACE \quad \text{(SAS)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q2 Answer: \(\displaystyle \Delta ABD \cong \Delta ACE\).

  3. Exercise 3

    CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that ΔADE≅ΔBCE\displaystyle \Delta \mathrm{ADE} \cong \Delta \mathrm{BCE}. NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-3_Q3

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    \[AD=BC \quad \text{(sides of square } ABCD\text{)} \] \[\angle ADC=\angle BCD=90^\circ \quad \text{(angles of a square)} \] \[\angle CDE=\angle DCE=60^\circ \quad \text{(angles of equilateral } \Delta CDE\text{)} \] \[\angle ADE=\angle ADC+\angle CDE=150^\circ,\ \angle BCE=\angle BCD+\angle DCE=150^\circ \] \[\Rightarrow \angle ADE=\angle BCE \] In \(\displaystyle \Delta ADE\) and \(\displaystyle \Delta BCE\): \[AD=BC,\ \angle ADE=\angle BCE,\ DE=CE\ (\text{sides of equilateral } \Delta CDE) \] \[\Rightarrow \Delta ADE \cong \Delta BCE \quad \text{(SAS)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q3 Answer: \(\displaystyle \Delta ADE \cong \Delta BCE\).

  4. Exercise 4

    In Fig. 7.6\displaystyle 7.6, BA⊥AC,DE⊥DF\displaystyle \mathrm{BA} \perp \mathrm{AC}, \mathrm{DE} \perp \mathrm{DF} such that BA=DE\displaystyle \mathrm{BA}=\mathrm{DE} and BF=EC\displaystyle \mathrm{BF}=\mathrm{EC}. Show that ΔABC≅ΔDEF\displaystyle \Delta \mathrm{ABC} \cong \Delta \mathrm{DEF}. NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-3_Q4

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    \[BF=EC \quad \text{(given)} \] \[BF+FC=EC+FC \Rightarrow BC=FE \quad \text{(adding common } FC\text{)} \] In right triangles ABC and DEF: \[\angle A=\angle D=90^\circ,\ BA=DE,\ BC=FE \] \[\Rightarrow \Delta ABC \cong \Delta DEF \quad \text{(RHS)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q4 Answer: \(\displaystyle \Delta ABC \cong \Delta DEF\).

  5. Exercise 5

    Q is a point on the side SR of a Δ\displaystyle \Delta PSR such that PQ = PR. Prove that PS > PQ.

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    \[PQ=PR \Rightarrow \angle PQR=\angle PRQ \quad \text{(isosceles } \Delta PQR\text{)} \] \[\angle PQS=180^\circ-\angle PQR \quad \text{(linear pair, } S,Q,R \text{ collinear)} \] \[\angle PSQ+\angle PRQ=180^\circ-\angle SPR<180^\circ \quad \text{(angle sum, } \Delta PSR\text{)} \] \[\Rightarrow \angle PSQ<180^\circ-\angle PRQ=\angle PQS \] \[\Rightarrow PS>PQ \quad \text{(side opposite greater angle, in } \Delta PSQ\text{)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q5 Answer: PS > PQ.

  6. Exercise 6

    S is any point on side QR of a ΔPQR\displaystyle \Delta \mathrm{PQR}. Show that: PQ+QR+RP>2PS\displaystyle \mathrm{PQ}+\mathrm{QR}+\mathrm{RP}>2 \mathrm{PS}.

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    In \(\displaystyle \Delta PQS\): \[PQ+QS>PS \quad \text{(sum of two sides of a triangle)} \] In \(\displaystyle \Delta PRS\): \[PR+RS>PS \] Adding, with \(\displaystyle QS+RS=QR\): \[PQ+PR+QR>2PS \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q6 Answer: PQ + QR + RP > 2PS.

  7. Exercise 7

    D is any point on side AC of a ΔABC\displaystyle \Delta \mathrm{ABC} with AB=AC\displaystyle \mathrm{AB}=\mathrm{AC}. Show that CD<BD\displaystyle \mathrm{CD}<\mathrm{BD}.

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    \[AB=AC \Rightarrow \angle ABC=\angle ACB \quad \text{(isosceles } \Delta ABC\text{)} \] \[\angle DBC=\angle ABC-\angle ABD<\angle ABC \quad \text{(BD lies inside } \angle ABC\text{)} \] \[\Rightarrow \angle DBC<\angle ACB=\angle BCD \] \[\Rightarrow CD<BD \quad \text{(side opposite smaller angle, in } \Delta BDC\text{)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q7 Answer: CD < BD.

  8. Exercise 8

    In Fig. 7.7\displaystyle 7.7, l∥m\displaystyle l \| m and M is the mid-point of a line segment AB. Show that M is also the mid-point of any line segment CD, having its end points on l\displaystyle l and m\displaystyle m, respectively. NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-3_Q8

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    \[\angle CAM=\angle DBM \quad \text{(alternate angles, } l\parallel m\text{)} \] \[AM=BM \quad \text{(given)} \] \[\angle AMC=\angle BMD \quad \text{(vertically opposite angles)} \] \[\Rightarrow \Delta AMC \cong \Delta BMD \quad \text{(ASA)} \] \[\Rightarrow CM=MD \quad \text{(CPCT)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q8 Answer: M is the midpoint of CD.

  9. Exercise 9

    Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. BO is produced to a point M . Prove that ∠MOC=\displaystyle \angle \mathrm{MOC}= ∠ABC\displaystyle \angle \mathrm{ABC}.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q9 \[\angle ABC = \angle ACB \quad \text{(angles opposite equal sides } AB=AC\text{)} \] Let \(\displaystyle \angle ABC = \angle ACB = 2\beta\); BO, CO bisect them, so \[\angle OBC = \angle OCB = \beta \] In \(\displaystyle \triangle OBC\), \[\angle BOC = 180^\circ - (\angle OBC + \angle OCB) = 180^\circ - 2\beta \quad \text{(angle sum property)} \] Since \(\displaystyle B, O, M\) are collinear, \[\angle MOC = 180^\circ - \angle BOC = 180^\circ - (180^\circ - 2\beta) = 2\beta \] Answer: \(\displaystyle \angle MOC = 2\beta = \angle ABC\).

  10. Exercise 10

    Bisectors of the angles B\displaystyle B and C\displaystyle C of an isosceles triangle ABC\displaystyle A B C with AB=AC\displaystyle A B=A C intersect each other at O . Show that external angle adjacent to ∠ABC\displaystyle \angle \mathrm{ABC} is equal to ∠BOC\displaystyle \angle \mathrm{BOC}.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q10 \[\angle ABC = \angle ACB \quad \text{(angles opposite equal sides } AB=AC\text{)} \] Let \(\displaystyle \angle ABC = \angle ACB = 2\beta\); BO, CO bisect them, so \[\angle OBC = \angle OCB = \beta \] In \(\displaystyle \triangle OBC\), \[\angle BOC = 180^\circ - (\angle OBC + \angle OCB) = 180^\circ - 2\beta \quad \text{(angle sum property)} \] Produce \(\displaystyle CB\) to \(\displaystyle E\); since \(\displaystyle C, B, E\) are collinear, \[\angle ABE = 180^\circ - \angle ABC = 180^\circ - 2\beta \quad \text{(linear pair)} \] Answer: \(\displaystyle \angle ABE = \angle BOC = 180^\circ - 2\beta\).

  11. Exercise 11

    In Fig. 7.8\displaystyle 7.8, AD is the bisector of ∠BAC\displaystyle \angle \mathrm{BAC}. Prove that AB>BD\displaystyle \mathrm{AB}>\mathrm{BD}. (E) Long Answer Questions Sample Question 1\displaystyle 1: In Fig. 7.9\displaystyle 7.9, ABC is a right triangle and right angled at B such that ∠BCA=2∠BAC\displaystyle \angle \mathrm{BCA}=2 \angle \mathrm{BAC}. Show that hypotenuse AC=2BC\displaystyle \mathrm{AC}=2 \mathrm{BC}. Solution: Produce CB to a point D such that BC=BD\displaystyle \mathrm{BC}=\mathrm{BD} and join AD. In ΔABC\displaystyle \Delta \mathrm{ABC} and ΔABD\displaystyle \Delta \mathrm{ABD}, we have BC=BD( By construction )AB=AB( Same side )∠ABC=∠ABD( Each of 90∘)\begin{aligned} \mathrm{BC} & =\mathrm{BD} & & (\text { By construction }) \\ \mathrm{AB} & =\mathrm{AB} & & (\text { Same side }) \\ \angle \mathrm{ABC} & =\angle \mathrm{ABD} & & \left(\text { Each of } 90^{\circ}\right) \end{aligned} Therefore, ΔABC≅ΔABD(SAS)\displaystyle \Delta \mathrm{ABC} \cong \Delta \mathrm{ABD} \quad(\mathrm{SAS})  So, ∠CAB=∠DAB and AC=AD Thus, ∠CAD=∠CAB+∠BAD=x+x=2x[ From (1) ] and ∠ACD=∠ADB=2x[ From (2),AC=AD](3)\begin{array}{rlrl} & \text { So, } & \angle \mathrm{CAB} & =\angle \mathrm{DAB} \\ \text { and } & \mathrm{AC} & =\mathrm{AD} \tag{3}\\ \text { Thus, } & \angle \mathrm{CAD} & =\angle \mathrm{CAB}+\angle \mathrm{BAD}=x+x=2 x \quad[\text { From (1) }] \\ \text { and } & \angle \mathrm{ACD} & =\angle \mathrm{ADB}=2 x \quad[\text { From }(2), \mathrm{AC}=\mathrm{AD}] \end{array} That is, ΔACD\displaystyle \Delta \mathrm{ACD} is an equilateral triangle. [From (3\displaystyle 3) and (4\displaystyle 4)] or AC=CD, i.e., AC=2BC (Since BC=BD ) \mathrm{AC}=\mathrm{CD} \text {, i.e., } \mathrm{AC}=2 \mathrm{BC} \text { (Since } \mathrm{BC}=\mathrm{BD} \text { ) } Sample Question 2\displaystyle 2 : Prove that if in two triangles two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent. Solution: See proof of Theorem 7.1\displaystyle 7.1 of Class IX Mathematics Textbook. Sample Question 3\displaystyle 3 : If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles. Solution : We are given a point D on side BC of a ΔABC\displaystyle \Delta \mathrm{ABC} such that ∠BAD=∠CAD\displaystyle \angle \mathrm{BAD}=\angle \mathrm{CAD} and BD=CD\displaystyle \mathrm{BD}=\mathrm{CD} (see Fig. 7.10\displaystyle 7.10). We are to prove that AB=AC\displaystyle \mathrm{AB}=\mathrm{AC}. Produce AD to a point E such that AD=DE\displaystyle \mathrm{AD}=\mathrm{DE} and then join CE . Now, in ΔABD\displaystyle \Delta \mathrm{ABD} and ΔECD\displaystyle \Delta \mathrm{ECD}, we have
    \begin{tabular}[t]{l} BD = CD
    (Given) AD=ED \mathrm{AD}=\mathrm{ED}
    (By construction)
    \hline and & ∠ADB=∠EDC\displaystyle \angle \mathrm{ADB}=\angle \mathrm{EDC} (Vertically opposite angles) \hline Therefore, & ΔABD≅ΔECD\displaystyle \Delta \mathrm{ABD} \cong \Delta \mathrm{ECD} (SAS) \hline So, and & AB=EC∠BAD=∠CED}(CPCT)(1)\left.\begin{array}{rl} \mathrm{AB} & =\mathrm{EC} \tag{1}\\ \angle \mathrm{BAD} & =\angle \mathrm{CED} \end{array}\right\}(\mathrm{CPCT}) \hline Also, & ∠BAD=∠CAD\displaystyle \angle \mathrm{BAD}=\angle \mathrm{CAD} (Given) \hline Therefore, & ∠CAD=∠CED[\displaystyle \angle \mathrm{CAD}=\angle \mathrm{CED} \quad[ From (2\displaystyle 2)] \hline So, & AC=EC\displaystyle \mathrm{AC}=\mathrm{EC} \hline Therefore, & AB = AC \hline \end{tabular} Sample Question 4\displaystyle 4 : S is any point in the interior of ΔPQR\displaystyle \Delta \mathrm{PQR}. Show that SQ+SR<\displaystyle \mathrm{SQ}+\mathrm{SR}< PQ + PR. Solution : Produce QS to intersect PR at T (See Fig. 7.11\displaystyle 7.11). From ΔPQT\displaystyle \Delta \mathrm{PQT}, we have PQ+PT>QT\displaystyle \mathrm{PQ}+\mathrm{PT}>\mathrm{QT} (Sum of any two sides is greater than the third side) PQ+PT>SQ+ST\begin{equation*} \mathrm{PQ}+\mathrm{PT}>\mathrm{SQ}+\mathrm{ST} \tag{1} \end{equation*} From ΔTSR\displaystyle \Delta \mathrm{TSR}, we have ST+TR>SR\begin{equation*} \mathrm{ST}+\mathrm{TR}>\mathrm{SR} \tag{2} \end{equation*} Adding (1\displaystyle 1) and (2\displaystyle 2), we get PQ+PT+ST+TR>SQ+ST+SR\mathrm{PQ}+\mathrm{PT}+\mathrm{ST}+\mathrm{TR}>\mathrm{SQ}+\mathrm{ST}+\mathrm{SR} i.e., PQ+PT+TR>SQ+SR\displaystyle \mathrm{PQ}+\mathrm{PT}+\mathrm{TR}>\mathrm{SQ}+\mathrm{SR} i.e., PQ+PR>SQ+SR\displaystyle \mathrm{PQ}+\mathrm{PR}>\mathrm{SQ}+\mathrm{SR} or SQ+SR<PQ+PR\displaystyle \mathrm{SQ}+\mathrm{SR}<\mathrm{PQ}+\mathrm{PR} NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-3_Q11 NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-3_Q11 NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-3_Q11 NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-3_Q11

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    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-3_Q11 Since \(\displaystyle B, D, C\) are collinear, \(\displaystyle \angle ADB\) is the exterior angle of \(\displaystyle \triangle ADC\) at \(\displaystyle D\): \[\angle ADB = \angle DAC + \angle ACD \quad \text{(exterior angle theorem)} \] \[\angle ADB > \angle DAC \] Since \(\displaystyle AD\) bisects \(\displaystyle \angle BAC\), \[\angle DAC = \angle BAD \] so \[\angle ADB > \angle BAD \] In \(\displaystyle \triangle ABD\), \(\displaystyle AB\) and \(\displaystyle BD\) lie opposite \(\displaystyle \angle ADB\) and \(\displaystyle \angle BAD\); the greater angle has the longer opposite side: \[AB > BD \] Answer: \(\displaystyle AB > BD\).