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Triangles

    Exercise 7.2

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9th class > Triangles > Exercise 7.2

Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that:

(i) OB = OC

(ii) AO bisects ∠ A

Instructions

Prove the above statement

  • Given: AB = AC and the angle bisectors of ∠B i.e. and ∠C i.e. intersect each other at point O.
  • ABC is an isosceles triangle with AB = AC, thus: ∠B = ∠
  • 12∠B = 12∠C i.e. ∠ = ∠
  • Since, ∠OBC = ∠OCB i.e. =
  • Considering the triangles ΔAOB and ΔAOC, AB = AC (given), AO = AO (common arm) and OB = OC (proven).
  • By congruency : ΔAOB ≅ Δ
  • By CPCT: ∠BAO = ∠ i.e. AO bisects ∠A.
  • Hence, proved.

In ∆ ABC, AD is the perpendicular bisector of BC. Show that ∆ ABC is an isosceles triangle in which AB = AC.

Solution:

Instructions

Considering ΔADB and ΔADC: AD = ( side)
∠ADB = ∠ (base angles in an triangle) and BD = (since is the perpendicular bisector)
So, ΔADB ≅ Δ by congruency criterion rule.
Thus, by CPST: AB =
Hence, proved.

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

Instructions

Prove the above statement

  • Considering the triangles ΔAEB and ΔAFC: ∠A = ∠A (common angle), ∠AEB = ∠AFC (as BE and CF are ), AB = AC (given)
  • By congruency: Δ ≅ ΔAFC
  • By CPCT: BE =
  • Hence, proved.

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that:

(i) ∆ ABE ≅ ∆ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Solution:

Instructions

Given: BE =
(i) Considering ΔABE and ΔACF: ∠AEB = ∠ ( angles)
∠A = ∠ ( angle) and BE = (Given)
Thus, ΔABE ≅ Δ by congruency condition.
By CPCT: AB = . _{span.reveal(when="blank-8")}So, ABC is an triangle.

5.ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ ABD = ∠ ACD.

We know that ABC and DBC are two isosceles triangles. We can draw the diagonal AD to form the triangles ΔABD and ΔACD.

Instructions

Prove the above statement

  • Considering the triangles ΔABD and ΔACD:AD = AD (common arm), AB = AC (ABC is an triangle) and BD = (BCD is an isosceles triangle)
  • By congruency rule: ΔABD ≅ Δ
  • Therefore, by CPCT: ∠ABD = ∠
  • Hence, proved.

6. ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB as shown below. Show that ∠ BCD is a right angle.

Instructions

Prove the above statement

  • Given that AB = AC and AD = AB. Thus, AB = = .This means that the triangles ΔABC and ΔACD are triangles.
  • Considering the triangle ΔABC: AB = AC and ∠ACB = ∠ as they are angles opposite to the sides.
  • Similarly, consider the triangle ΔACD: AD = which makes ∠ADC = ∠
  • In the case of ΔABC, ∠CAB + ∠ACB + ∠ABC = °
  • Substituting values
  • To get the equation (1)
  • Similarly,in the triangle ΔADC: ∠CAD = 180° – ∠ACD — (2)
  • From the figure that:∠CAB + ∠CAD = 180° (as BD is a line)
  • Adding the equations (1) and (2) we get
  • Simplifying: (∠ACB + ∠ACD) = 180°
  • Which gives: ∠ = °
  • Hence, proved.

ABC is a right angled triangle in which ∠ A = 90° and AB = AC. Find ∠ B and ∠ C.

Instructions

Given: ∠A = ° and AB = ⇒ ∠B = ∠ (They are angles to the equal sides, so they are )
We have: ∠A + ∠B + ∠C = ° (sum of the interior angles of the triangle)
Thus, ° + ∠B = ° ⇒ 2∠B = ° ⇒ ∠B = °
Thus, ∠B = ∠C = °

8.Show that the angles of an equilateral triangle are 60° each.

Instructions

Prove the above statement

  • Let ABC be an equilateral triangle as shown above.Thus, AB = = (enter sides in anticlockwise direction)
  • We also know that: ∠A+ ∠B + ∠C = 180°. Thus, ∠A = 180° i.e. ∠A = °
  • Therefore, ∠A = ∠B = ∠C = °
  • Thus, all angles of an equilateral triangle are always 60°.