Triangles | Exercise 6.3

Question 7

  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}

Question diagram 1
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Solution

We will use the following similarity criterion:

  • AA (Angle-Angle): If two angles of one triangle are equal to two corresponding angles of another triangle, the triangles are similar.

We will use the Angle-Angle (AA) similarity criterion to prove each part.

Step 1 — Proving ΔAEP∼ΔCDP\Delta \text{AEP} \sim \Delta \text{CDP}

Let's look at ΔAEP\Delta AEP and ΔCDP\Delta CDP. We know CE is an altitude. So, ∠AEP=90∘\angle AEP = 90^\circ. We know AD is an altitude. So, ∠CDP=90∘\angle CDP = 90^\circ. Thus, ∠AEP=∠CDP\angle AEP = \angle CDP. These are right angles. ∠APE\angle APE and ∠CPD\angle CPD are vertically opposite angles. So, ∠APE=∠CPD\angle APE = \angle CPD. By AA similarity, the triangles are similar.

ΔAEP∼ΔCDP\boxed{\Delta \text{AEP} \sim \Delta \text{CDP}}

Diagram 1

Step 2 — Proving ΔABD∼ΔCBE\Delta \text{ABD} \sim \Delta \text{CBE}

Let's look at ΔABD\Delta ABD and ΔCBE\Delta CBE. We know AD is an altitude. So, ∠ADB=90∘\angle ADB = 90^\circ. We know CE is an altitude. So, ∠CEB=90∘\angle CEB = 90^\circ. Thus, ∠ADB=∠CEB\angle ADB = \angle CEB. These are right angles. ∠B\angle B is common to both triangles. So, ∠ABD=∠CBE\angle ABD = \angle CBE. By AA similarity, the triangles are similar.

ΔABD∼ΔCBE\boxed{\Delta \text{ABD} \sim \Delta \text{CBE}}

Diagram 2

Step 3 — Proving ΔAEP∼ΔADB\Delta \text{AEP} \sim \Delta \text{ADB}

Let's look at ΔAEP\Delta AEP and ΔADB\Delta ADB. We know CE is an altitude. So, ∠AEP=90∘\angle AEP = 90^\circ. We know AD is an altitude. So, ∠ADB=90∘\angle ADB = 90^\circ. Thus, ∠AEP=∠ADB\angle AEP = \angle ADB. These are right angles. ∠A\angle A is common to both triangles. So, ∠PAE=∠DAB\angle PAE = \angle DAB. By AA similarity, the triangles are similar.

ΔAEP∼ΔADB\boxed{\Delta \text{AEP} \sim \Delta \text{ADB}}

Diagram 3

Step 4 — Proving ΔPDC∼ΔBEC\Delta \text{PDC} \sim \Delta \text{BEC}

Let's look at ΔPDC\Delta PDC and ΔBEC\Delta BEC. We know AD is an altitude. So, ∠PDC=90∘\angle PDC = 90^\circ. We know CE is an altitude. So, ∠BEC=90∘\angle BEC = 90^\circ. Thus, ∠PDC=∠BEC\angle PDC = \angle BEC. These are right angles. ∠C\angle C is common to both triangles. So, ∠PCD=∠BCE\angle PCD = \angle BCE. By AA similarity, the triangles are similar.

ΔPDC∼ΔBEC\boxed{\Delta \text{PDC} \sim \Delta \text{BEC}}

Answer

(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}

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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