Triangles | Exercise 6.3

Question 12

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

Question diagram 1
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Solution

We will use the following similarity criterion:

  • SSS (Side-Side-Side): If the ratios of all three corresponding sides of two triangles are equal, the triangles are similar.
  • SAS (Side-Angle-Side): If two sides of one triangle are proportional to two corresponding sides of another triangle, and the included angles are equal, the triangles are similar.

We need to show that triangle ABC is similar to triangle PQR.

Step 1 — Understand Medians

AD is a median of Δ\Delta ABC. This means D is the midpoint of BC.

So, we can write:

BD=12BCBD = \frac{1}{2} BC

PM is a median of Δ\Delta PQR. This means M is the midpoint of QR.

So, we can write:

QM=12QRQM = \frac{1}{2} QR

Diagram 1

Step 2 — Adjust Proportionality

We are given that the sides and medians are proportional.

ABPQ=BCQR=ADPM\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}

Let's substitute the expressions for BC and QR from Step 1.

ABPQ=2BD2QM=ADPM\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM}

We can simplify this expression.

ABPQ=BDQM=ADPM\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}

ABPQ=BDQM=ADPM\boxed{\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}}

Step 3 — Prove Similarity of Smaller Triangles

Now, let's consider Δ\Delta ABD and Δ\Delta PQM.

From Step 2, we have shown that their corresponding sides are proportional.

ABPQ=BDQM=ADPM\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}

Therefore, by the SSS (Side-Side-Side) similarity criterion, Δ\Delta ABD is similar to Δ\Delta PQM.

ΔABD∼ΔPQM\boxed{\Delta ABD \sim \Delta PQM}

Step 4 — Find Equal Angles

Since Δ\Delta ABD ∼\sim Δ\Delta PQM, their corresponding angles must be equal.

So, we can say:

∠ABD=∠PQM\angle ABD = \angle PQM

Note that ∠\angle ABD is the same as ∠\angle ABC. Also, ∠\angle PQM is the same as ∠\angle PQR.

Therefore, we have:

∠ABC=∠PQR\angle ABC = \angle PQR

∠ABC=∠PQR\boxed{\angle ABC = \angle PQR}

Step 5 — Prove Similarity of Larger Triangles

Now, let's consider Δ\Delta ABC and Δ\Delta PQR.

We are given that two sides are proportional.

ABPQ=BCQR\frac{AB}{PQ} = \frac{BC}{QR}

From Step 4, we have shown that the included angles are equal.

∠ABC=∠PQR\angle ABC = \angle PQR

Therefore, by the SAS (Side-Angle-Side) similarity criterion, Δ\Delta ABC is similar to Δ\Delta PQR.

ΔABC∼ΔPQR\boxed{\Delta ABC \sim \Delta PQR}

Answer

(i) We used the definition of a median to relate the base segments to the full base. (ii) We proved Δ\Delta ABD ∼\sim Δ\Delta PQM using SSS similarity. (iii) We then used SAS similarity to show Δ\Delta ABC ∼\sim Δ\Delta PQR.

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 and BD of a trapezium ABCD with AB∥DC\text{AB} \parallel \text{DC} intersect each other at the point O. Using a similarity criterion for two triangles, show that OAOC=OBOD\frac{\text{OA}}{\text{OC}} = \frac{\text{OB}}{\text{OD}}.
Q4
  1. In Fig. 6.36, QRQS=QTPR\frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}} and ∠1=∠2\angle 1 = \angle 2. Show that ΔPQS∼ΔTQR\Delta \text{PQS} \sim \Delta \text{TQR}.
Q5
  1. S and T are points on sides PR and QR 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 AD and CE of ΔABC\Delta \text{ABC} intersect each other at the point P. 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\frac{\text{CA}}{\text{PA}} = \frac{\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\frac{\text{CD}}{\text{GH}} = \frac{\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, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥\perp BC and EF ⊥\perp AC, prove that Δ\Delta ABD ∼\sim Δ\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 Δ\Delta PQR (see Fig. 6.41). Show that Δ\Delta ABC ∼\sim Δ\Delta PQR.

Q13

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

Q14

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that Δ\Delta ABC ∼\sim Δ\Delta PQR.

Q15

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Q16

If AD and PM are medians of triangles ABC and PQR, respectively where Δ\Delta ABC ∼\sim Δ\Delta PQR, prove that

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

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