Triangles | Exercise 6.3

Question 6

  1. In Fig. 6.37, if ΔABEΔACD\Delta \text{ABE} \cong \Delta \text{ACD}, show that ΔADEΔABC\Delta \text{ADE} \sim \Delta \text{ABC}.
Question diagram 1
Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • Given that ΔABEΔACD\Delta \text{ABE} \cong \Delta \text{ACD}, their corresponding parts are equal (CPCT):
    • AB=AC\text{AB} = \text{AC}
    • AD=AE\text{AD} = \text{AE}
  • Dividing these relations gives the proportional ratio of sides: ADAB=AEAC\dfrac{\text{AD}}{\text{AB}} = \dfrac{\text{AE}}{\text{AC}}.
  • Both ΔADE\Delta \text{ADE} and ΔABC\Delta \text{ABC} share the common angle A\angle \text{A}.
  • By the SAS (Side-Angle-Side) similarity criterion, if two sides are proportional and the included angle is equal, the triangles are similar: ΔADEΔABC\Delta \text{ADE} \sim \Delta \text{ABC}.

Step 1 · Identify Equal Corresponding Sides

Diagram 1

Given ΔABEΔACD\Delta \text{ABE} \cong \Delta \text{ACD}.

By Corresponding Parts of Congruent Triangles (CPCT) AB=AC(1)\text{AB} = \text{AC} \quad \dots (1) AD=AE(2)\text{AD} = \text{AE} \quad \dots (2)

Step 2 · Form the Ratio of Sides

Dividing equation (2)(2) by equation (1)(1) ADAB=AEAC(3)\dfrac{\text{AD}}{\text{AB}} = \dfrac{\text{AE}}{\text{AC}} \quad \dots (3)

Step 3 · Apply SAS Similarity Criterion

In ΔADE\Delta \text{ADE} and ΔABC\Delta \text{ABC} ADAB=AEAC[From (3)]\dfrac{\text{AD}}{\text{AB}} = \dfrac{\text{AE}}{\text{AC}} \quad [\text{From }(3)]

DAE=BAC[Common angle]\angle \text{DAE} = \angle \text{BAC} \quad [\text{Common angle}]

Therefore, by SAS similarity criterion ΔADEΔABC\Delta \text{ADE} \sim \Delta \text{ABC}

Answer

ΔADEΔABC\Delta \text{ADE} \sim \Delta \text{ABC} (proved by SAS similarity)

Common Mistakes
  • Mismatched CPCT sides: Writing AB=AD\text{AB} = \text{AD} instead of matching corresponding vertices from ΔABEΔACD\Delta \text{ABE} \cong \Delta \text{ACD}, which gives AB=AC\text{AB} = \text{AC} and AE=AD\text{AE} = \text{AD}.
  • Confusing Congruence with Similarity: Congruence requires equal side lengths (ratio 1:11:1), whereas similarity only requires proportional sides (ADAB=AEAC)\left(\dfrac{\text{AD}}{\text{AB}} = \dfrac{\text{AE}}{\text{AC}}\right) and equal corresponding angles.

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, ΔODCΔOBA\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 ABDC\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 ΔPQSΔTQR\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 ΔRPQΔRTS\Delta \text{RPQ} \sim \Delta \text{RTS}.

Q6
  1. In Fig. 6.37, if ΔABEΔACD\Delta \text{ABE} \cong \Delta \text{ACD}, show that ΔADEΔABC\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) ΔAEPΔCDP\Delta \text{AEP} \sim \Delta \text{CDP}

(ii) ΔABDΔCBE\Delta \text{ABD} \sim \Delta \text{CBE}

(iii) ΔAEPΔADB\Delta \text{AEP} \sim \Delta \text{ADB}

(iv) ΔPDCΔBEC\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 ΔABEΔCFB\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) ΔABCΔAMP\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 ΔABCΔFEG\Delta \text{ABC} \sim \Delta \text{FEG}, show that:

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

(ii) ΔDCBΔHGE\Delta \text{DCB} \sim \Delta \text{HGE}

(iii) ΔDCAΔHGF\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 ADBCAD \perp BC and EFACEF \perp AC, prove that ΔABDΔECF\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 ΔABCΔPQR\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=CBCD\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 ΔABCΔPQR\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 ΔABCΔPQR\Delta ABC \sim \Delta PQR, prove that

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

← Back to Triangles