Triangles | Exercise 6.3

Question 7

  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}

Question diagram 1
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Solution
Understand the Question
  • In ΔABC\Delta \text{ABC}, AD⊥BCAD \perp BC and CE⊥ABCE \perp AB, so ∠ADC=∠ADB=90∘\angle ADC = \angle ADB = 90^\circ and ∠AEB=∠CEB=∠AEP=90∘\angle AEB = \angle CEB = \angle AEP = 90^\circ.
  • To prove the triangles are similar in each part, we use the AA (Angle-Angle) Similarity Criterion: if two angles of one triangle are equal to two corresponding angles of another triangle, then the two triangles are similar.

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

Step 1 · Prove Similarity using AA Criterion

Diagram 1

In ΔAEP\Delta \text{AEP} and ΔCDP\Delta \text{CDP}:

∠AEP=∠CDP=90∘(Since CE⊥AB and AD⊥BC)∠APE=∠CPD(Vertically opposite angles)\begin{aligned} \angle \text{AEP} &= \angle \text{CDP} = 90^\circ && (\text{Since } CE \perp AB \text{ and } AD \perp BC) \\[0.4em] \angle \text{APE} &= \angle \text{CPD} && (\text{Vertically opposite angles}) \end{aligned}

Therefore, by AA similarity criterion: ΔAEP∼ΔCDP\Delta \text{AEP} \sim \Delta \text{CDP}

Answer

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

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

Step 1 · Prove Similarity using AA Criterion

Diagram 2

In ΔABD\Delta \text{ABD} and ΔCBE\Delta \text{CBE}:

∠ADB=∠CEB=90∘(Since AD⊥BC and CE⊥AB)∠ABD=∠CBE(Common angle ∠B)\begin{aligned} \angle \text{ADB} &= \angle \text{CEB} = 90^\circ && (\text{Since } AD \perp BC \text{ and } CE \perp AB) \\[0.4em] \angle \text{ABD} &= \angle \text{CBE} && (\text{Common angle } \angle B) \end{aligned}

Therefore, by AA similarity criterion: ΔABD∼ΔCBE\Delta \text{ABD} \sim \Delta \text{CBE}

Answer

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

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

Step 1 · Prove Similarity using AA Criterion

Diagram 3

In ΔAEP\Delta \text{AEP} and ΔADB\Delta \text{ADB}:

∠AEP=∠ADB=90∘(Since CE⊥AB and AD⊥BC)∠PAE=∠DAB(Common angle ∠A)\begin{aligned} \angle \text{AEP} &= \angle \text{ADB} = 90^\circ && (\text{Since } CE \perp AB \text{ and } AD \perp BC) \\[0.4em] \angle \text{PAE} &= \angle \text{DAB} && (\text{Common angle } \angle A) \end{aligned}

Therefore, by AA similarity criterion: ΔAEP∼ΔADB\Delta \text{AEP} \sim \Delta \text{ADB}

Answer

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

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

Step 1 · Prove Similarity using AA Criterion

In ΔPDC\Delta \text{PDC} and ΔBEC\Delta \text{BEC}:

∠PDC=∠BEC=90∘(Since AD⊥BC and CE⊥AB)∠PCD=∠BCE(Common angle ∠C)\begin{aligned} \angle \text{PDC} &= \angle \text{BEC} = 90^\circ && (\text{Since } AD \perp BC \text{ and } CE \perp AB) \\[0.4em] \angle \text{PCD} &= \angle \text{BCE} && (\text{Common angle } \angle C) \end{aligned}

Therefore, by AA similarity criterion: ΔPDC∼ΔBEC\Delta \text{PDC} \sim \Delta \text{BEC}

Answer

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

Common Mistakes
  • Vertex Correspondence Error: Writing triangle similarity without maintaining the correct order of corresponding vertices (e.g., writing ΔAEP∼ΔDCP\Delta \text{AEP} \sim \Delta \text{DCP} instead of ΔCDP\Delta \text{CDP}).
  • Altitude Angle Confusion: Assuming an altitude bisects the angle instead of simply forming 90∘90^\circ with the opposite side.

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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