Triangles | Exercise 6.3

Question 9

  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}}

Question diagram 1
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Solution

We will use the AA similarity criterion to prove the triangles are similar, then use the property of similar triangles for side ratios.

Step 1 — Show triangle similarity

Let's look at the two triangles, ΔABC\Delta \text{ABC} and ΔAMP\Delta \text{AMP}. We are given that ΔABC\Delta \text{ABC} is right-angled at B. This means that ∠ABC\angle \text{ABC} is 90 degrees. We are also given that ΔAMP\Delta \text{AMP} is right-angled at M. This means that ∠AMP\angle \text{AMP} is 90 degrees. So, we have one pair of equal angles.

∠ABC=∠AMP\angle \text{ABC} = \angle \text{AMP}

=90∘= 90^\circ

Now, let's look at angle A. Angle A is part of ΔABC\Delta \text{ABC} (as ∠BAC\angle \text{BAC}). Angle A is also part of ΔAMP\Delta \text{AMP} (as ∠MAP\angle \text{MAP}). So, angle A is common to both triangles.

∠BAC=∠MAP\angle \text{BAC} = \angle \text{MAP}

=∠A (Common Angle)= \angle \text{A} \text{ (Common Angle)}

Since two angles of ΔABC\Delta \text{ABC} are equal to two angles of ΔAMP\Delta \text{AMP}, the triangles are similar by the AA similarity criterion.

ΔABC∼ΔAMP\boxed{\Delta \text{ABC} \sim \Delta \text{AMP}}

Diagram 1

Step 2 — Show ratio of sides

We have already proved that ΔABC\Delta \text{ABC} is similar to ΔAMP\Delta \text{AMP}. When two triangles are similar, the ratio of their corresponding sides is equal. Let's identify the corresponding vertices. Angle A corresponds to angle A. Angle B (90 degrees) corresponds to angle M (90 degrees). Angle C corresponds to angle P. So, the corresponding sides are AB and AM, BC and MP, CA and PA. We can write the ratios of these corresponding sides.

ABAM=BCMP=CAPA\frac{\text{AB}}{\text{AM}} = \frac{\text{BC}}{\text{MP}} = \frac{\text{CA}}{\text{PA}}

The question asks us to prove a specific part of this equality. We need to show that CAPA=BCMP\frac{\text{CA}}{\text{PA}} = \frac{\text{BC}}{\text{MP}}. This is directly obtained from the ratios of corresponding sides of similar triangles.

CAPA=BCMP\boxed{\frac{\text{CA}}{\text{PA}} = \frac{\text{BC}}{\text{MP}}}

Answer

(i) ΔABC∼ΔAMP\Delta \text{ABC} \sim \Delta \text{AMP} (ii) CAPA=BCMP\frac{\text{CA}}{\text{PA}} = \frac{\text{BC}}{\text{MP}}

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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