Triangles | Exercise 6.3

Question 16

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}}

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Solution
Understand the Question
  • Given that ΔABCΔPQR\Delta \text{ABC} \sim \Delta \text{PQR}, their corresponding sides are proportional (ABPQ=BCQR\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{BC}}{\text{QR}}) and their corresponding angles are equal (B=Q\angle \text{B} = \angle \text{Q}).
  • Since AD\text{AD} and PM\text{PM} are medians, they bisect the base sides such that BD=12BC\text{BD} = \dfrac{1}{2}\text{BC} and QM=12QR\text{QM} = \dfrac{1}{2}\text{QR}.
  • By replacing BC\text{BC} and QR\text{QR} with their halves in the ratio, we prove that the smaller triangles ΔABD\Delta \text{ABD} and ΔPQM\Delta \text{PQM} are similar by the SAS similarity criterion, which directly leads to ABPQ=ADPM\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{AD}}{\text{PM}}.

Step 1 · Express Side Ratios Using Medians

Diagram 1

Given ΔABCΔPQR\Delta \text{ABC} \sim \Delta \text{PQR}.

Corresponding sides are proportional: ABPQ=BCQR=ACPR(1)\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{BC}}{\text{QR}} = \dfrac{\text{AC}}{\text{PR}} \quad \dots (1)

Corresponding angles are equal:

A=PB=Q(2)C=R\begin{aligned} \angle \text{A} &= \angle \text{P} \\ \angle \text{B} &= \angle \text{Q} \quad \dots (2) \\ \angle \text{C} &= \angle \text{R} \end{aligned}

Since AD\text{AD} and PM\text{PM} are medians to BC\text{BC} and QR\text{QR} respectively: BD=BC2(3)\text{BD} = \dfrac{\text{BC}}{2} \quad \dots (3)

QM=QR2(4)\text{QM} = \dfrac{\text{QR}}{2} \quad \dots (4)

From (1)(1), (3)(3), and (4)(4): BCQR=ABPQ\dfrac{\text{BC}}{\text{QR}} = \dfrac{\text{AB}}{\text{PQ}}

2×BD2×QM=ABPQ\dfrac{2 \times \text{BD}}{2 \times \text{QM}} = \dfrac{\text{AB}}{\text{PQ}}

BDQM=ABPQ(5)\dfrac{\text{BD}}{\text{QM}} = \dfrac{\text{AB}}{\text{PQ}} \quad \dots (5)

Step 2 · Prove Similarity of ΔABD\Delta \text{ABD} and ΔPQM\Delta \text{PQM}

In ΔABD\Delta \text{ABD} and ΔPQM\Delta \text{PQM}:

From (2)(2): B=Q\angle \text{B} = \angle \text{Q}

From (5)(5): ABPQ=BDQM\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{BD}}{\text{QM}}

Therefore, by the SAS similarity criterion: ΔABDΔPQM\Delta \text{ABD} \sim \Delta \text{PQM}

Since corresponding sides of similar triangles are in proportion: ABPQ=BDQM=ADPM\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{BD}}{\text{QM}} = \dfrac{\text{AD}}{\text{PM}}

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

Answer

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

Common Mistakes
  • Median vs. Altitude: A median bisects the opposite side (BD=12BC\text{BD} = \frac{1}{2}\text{BC}), but does not necessarily meet it at a right angle (9090^\circ). Do not assume ADB=90\angle \text{ADB} = 90^\circ.
  • Incorrect Similarity Criterion: Trying to use AAA or SSS directly on ΔABD\Delta \text{ABD} and ΔPQM\Delta \text{PQM} before establishing the relationship BDQM=ABPQ\dfrac{\text{BD}}{\text{QM}} = \dfrac{\text{AB}}{\text{PQ}}, which is necessary for the SAS criterion.

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}}

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