Triangles | Exercise 6.3

Question 10

  1. CD and GH are respectively the bisectors of ∠ACB\angle \text{ACB} and ∠EGF\angle \text{EGF} such that D and H lie on sides AB and FE of ΔABC\Delta \text{ABC} and ΔEFG\Delta \text{EFG} respectively. If ΔABC∼ΔFEG\Delta \text{ABC} \sim \Delta \text{FEG}, show that:

(i) CDGH=ACFG\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}

(ii) ΔDCB∼ΔHGE\Delta \text{DCB} \sim \Delta \text{HGE}

(iii) ΔDCA∼ΔHGF\Delta \text{DCA} \sim \Delta \text{HGF}

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Solution
Understand the Question
  • Given that ΔABC∼ΔFEG\Delta \text{ABC} \sim \Delta \text{FEG}, corresponding angles are equal: ∠A=∠F,∠B=∠E,∠ACB=∠FGE\angle \text{A} = \angle \text{F}, \quad \angle \text{B} = \angle \text{E}, \quad \angle \text{ACB} = \angle \text{FGE}
  • CD\text{CD} and GH\text{GH} are the angle bisectors of ∠ACB\angle \text{ACB} and ∠EGF\angle \text{EGF} respectively: ∠ACD=∠BCD=12∠ACB\angle \text{ACD} = \angle \text{BCD} = \dfrac{1}{2}\angle \text{ACB} ∠FGH=∠EGH=12∠FGE\angle \text{FGH} = \angle \text{EGH} = \dfrac{1}{2}\angle \text{FGE}
  • Since ∠ACB=∠FGE\angle \text{ACB} = \angle \text{FGE}, their halves are equal: ∠ACD=∠FGHand∠BCD=∠EGH\angle \text{ACD} = \angle \text{FGH} \quad \text{and} \quad \angle \text{BCD} = \angle \text{EGH}
  • We use the AA similarity criterion to establish similarity between the required pairs of triangles.

(i) Show that CDGH=ACFG\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}

Step 1 · Prove Similarity of ΔDCA\Delta \text{DCA} and ΔHGF\Delta \text{HGF} to Find Ratio of Sides

Diagram 1

In ΔDCA\Delta \text{DCA} and ΔHGF\Delta \text{HGF}:

∠A=∠F(∵ΔABC∼ΔFEG)\angle \text{A} = \angle \text{F} \quad (\because \Delta \text{ABC} \sim \Delta \text{FEG})

∠ACD=∠FGH(∵12∠ACB=12∠FGE)\angle \text{ACD} = \angle \text{FGH} \quad \left(\because \dfrac{1}{2}\angle \text{ACB} = \dfrac{1}{2}\angle \text{FGE}\right)

By AA similarity criterion:

ΔDCA∼ΔHGF\Delta \text{DCA} \sim \Delta \text{HGF}

Since corresponding sides of similar triangles are proportional:

CDGH=ACFG=ADHF\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}} = \dfrac{\text{AD}}{\text{HF}}

  ⟹  CDGH=ACFG\implies \dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}

Answer

(i) CDGH=ACFG\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}

(ii) Show that ΔDCB∼ΔHGE\Delta \text{DCB} \sim \Delta \text{HGE}

Step 1 · Apply AA Similarity Criterion to ΔDCB\Delta \text{DCB} and ΔHGE\Delta \text{HGE}

In ΔDCB\Delta \text{DCB} and ΔHGE\Delta \text{HGE}:

∠B=∠E(∵ΔABC∼ΔFEG)\angle \text{B} = \angle \text{E} \quad (\because \Delta \text{ABC} \sim \Delta \text{FEG})

∠BCD=∠EGH(∵12∠ACB=12∠FGE)\angle \text{BCD} = \angle \text{EGH} \quad \left(\because \dfrac{1}{2}\angle \text{ACB} = \dfrac{1}{2}\angle \text{FGE}\right)

By AA similarity criterion:

ΔDCB∼ΔHGE\Delta \text{DCB} \sim \Delta \text{HGE}

Answer

(ii) ΔDCB∼ΔHGE\Delta \text{DCB} \sim \Delta \text{HGE}

(iii) Show that ΔDCA∼ΔHGF\Delta \text{DCA} \sim \Delta \text{HGF}

Step 1 · Apply AA Similarity Criterion to ΔDCA\Delta \text{DCA} and ΔHGF\Delta \text{HGF}

In ΔDCA\Delta \text{DCA} and ΔHGF\Delta \text{HGF}:

∠A=∠F(∵ΔABC∼ΔFEG)\angle \text{A} = \angle \text{F} \quad (\because \Delta \text{ABC} \sim \Delta \text{FEG})

∠ACD=∠FGH(∵12∠ACB=12∠FGE)\angle \text{ACD} = \angle \text{FGH} \quad \left(\because \dfrac{1}{2}\angle \text{ACB} = \dfrac{1}{2}\angle \text{FGE}\right)

By AA similarity criterion:

ΔDCA∼ΔHGF\Delta \text{DCA} \sim \Delta \text{HGF}

Answer

(iii) ΔDCA∼ΔHGF\Delta \text{DCA} \sim \Delta \text{HGF}

Common Mistakes
  • Vertex Correspondence Error: Given ΔABC∼ΔFEG\Delta \text{ABC} \sim \Delta \text{FEG}, vertex A\text{A} corresponds to F\text{F}, B\text{B} to E\text{E}, and C\text{C} to G\text{G}. Do not assume conventional alphabetical pairing like A↔E\text{A} \leftrightarrow \text{E} or B↔F\text{B} \leftrightarrow \text{F}.
  • Proof Dependency: Part (i) directly depends on the similarity proved in part (iii). Proving ΔDCA∼ΔHGF\Delta \text{DCA} \sim \Delta \text{HGF} first is required before equating the side ratios.

More questions in Exercise 6.3

Q1
  1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Q2
  1. In Fig. 6.35, ΔODC∼ΔOBA\Delta \text{ODC} \sim \Delta \text{OBA}, ∠BOC=125∘\angle \text{BOC} = 125^\circ and ∠CDO=70∘\angle \text{CDO} = 70^\circ. Find ∠DOC\angle \text{DOC}, ∠DCO\angle \text{DCO} and ∠OAB\angle \text{OAB}.
Q3
  1. Diagonals AC\text{AC} and BD\text{BD} of a trapezium ABCD\text{ABCD} with AB∥DC\text{AB} \parallel \text{DC} intersect each other at the point O\text{O}. Using a similarity criterion for two triangles, show that OAOC=OBOD\dfrac{\text{OA}}{\text{OC}} = \dfrac{\text{OB}}{\text{OD}}.
Q4
  1. In Fig. 6.36, QRQS=QTPR\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}} and ∠1=∠2\angle 1 = \angle 2. Show that ΔPQS∼ΔTQR\Delta \text{PQS} \sim \Delta \text{TQR}.
Q5

S and T are points on sides PRPR and QRQR of ΔPQR\Delta \text{PQR} such that ∠P=∠RTS\angle \text{P} = \angle \text{RTS}. Show that ΔRPQ∼ΔRTS\Delta \text{RPQ} \sim \Delta \text{RTS}.

Q6
  1. In Fig. 6.37, if ΔABE≅ΔACD\Delta \text{ABE} \cong \Delta \text{ACD}, show that ΔADE∼ΔABC\Delta \text{ADE} \sim \Delta \text{ABC}.
Q7
  1. In Fig. 6.38, altitudes ADAD and CECE of ΔABC\Delta \text{ABC} intersect each other at the point PP. Show that:

(i) ΔAEP∼ΔCDP\Delta \text{AEP} \sim \Delta \text{CDP}

(ii) ΔABD∼ΔCBE\Delta \text{ABD} \sim \Delta \text{CBE}

(iii) ΔAEP∼ΔADB\Delta \text{AEP} \sim \Delta \text{ADB}

(iv) ΔPDC∼ΔBEC\Delta \text{PDC} \sim \Delta \text{BEC}

Q8
  1. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE∼ΔCFB\Delta \text{ABE} \sim \Delta \text{CFB}.
Q9
  1. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) ΔABC∼ΔAMP\Delta \text{ABC} \sim \Delta \text{AMP}

(ii) CAPA=BCMP\dfrac{\text{CA}}{\text{PA}} = \dfrac{\text{BC}}{\text{MP}}

Q10
  1. CD and GH are respectively the bisectors of ∠ACB\angle \text{ACB} and ∠EGF\angle \text{EGF} such that D and H lie on sides AB and FE of ΔABC\Delta \text{ABC} and ΔEFG\Delta \text{EFG} respectively. If ΔABC∼ΔFEG\Delta \text{ABC} \sim \Delta \text{FEG}, show that:

(i) CDGH=ACFG\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}

(ii) ΔDCB∼ΔHGE\Delta \text{DCB} \sim \Delta \text{HGE}

(iii) ΔDCA∼ΔHGF\Delta \text{DCA} \sim \Delta \text{HGF}

Q11

In Fig. 6.40, EE is a point on side CBCB produced of an isosceles triangle ABCABC with AB=ACAB = AC. If AD⊥BCAD \perp BC and EF⊥ACEF \perp AC, prove that ΔABD∼ΔECF\Delta ABD \sim \Delta ECF.

Q12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR\Delta PQR (see Fig. 6.41). Show that ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR.

Q13

D is a point on the side BCBC of a triangle ABCABC such that ∠ADC=∠BAC\angle \text{ADC} = \angle \text{BAC}. Show that CA2=CB⋅CD\text{CA}^2 = \text{CB} \cdot \text{CD}.

Q14

Sides ABAB and ACAC and median ADAD of a triangle ABCABC are respectively proportional to sides PQPQ and PRPR and median PMPM of another triangle PQRPQR. Show that ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR.

Q15

A vertical pole of length 6 m6\text{ m} casts a shadow 4 m4\text{ m} long on the ground and at the same time a tower casts a shadow 28 m28\text{ m} long. Find the height of the tower.

Q16

If ADAD and PMPM are medians of triangles ABCABC and PQRPQR, respectively where ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR, prove that

ABPQ=ADPM\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{AD}}{\text{PM}}

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