7.3 SOME PROPERTIES Of A TRIANGLE:
* Sides opposite to equal angles are equal.
* Angles opposite to equal sides are equal.
* A triangle in which two sides are equal is called an isosceles triangle.
* Two figures are congruent it they are of the same shape and size.
* In a triangle the angle opposite to the larger side is greater.
* Sum of two sides of a triangle is always greater than the third side.
* Two angles are congruent only when their measures are equal.
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.
Given: AB = AC and the bisectors of ∠B and ∠A meet at O
To prove: (i) OB = OC (ii) AO bisects ∠A
Proof: (i) AB = AC (given)
∴ ∴ ∠C = ∠B (angles opposite to equal sides are equal )
∴ 1/2 ∠C = 1/2 ∠B
∴ ∠OCB = ∠OBC
∴ OB = OC (sides opposite to equal angles are equal)
(ii) In Δ ABO and Δ ACO we have
AB = AC (given)
OB = OC (proved)
∠OBA = ∠OCA (1/2∠B = 1/2∠C)
By S.A.S congruence rule Δ ABO ≅ Δ ACO
∴ ∠OAB = ∠OAC (c.p.c.t)
∴ AO bisects ∠A
2. In Δ ABC, AD is the perpendicular bisector of BC . Show that Δ ABC is an isosceles triangle in which AB = AC.
Given: AD is perpendicular bisector means BD = CD and
∠ADB = ∠ADC = 90°
To prove: AB = AC
Proof: In Δ ABD and Δ ACD we have
AD = AD (common side)
∠ADB = ∠ADC (given)
BD = CD (given)
By S.A.S congruence rule
⊿ABD ≅⊿ACD
∴ Their corresponding parts are equal
∴ AB = AC (c.p.c.t)
∴ ΔABC is an isosceles triangle.
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.
Given: ABC is an isosceles triangle i.e: AB = AC
To prove: BE = CF
Proof: Since AB = AC (given)
∠ACB = ∠ABC (Angles opposite to equal sides are equal)
Now in Δ BEC and Δ CFB we have
∠EBC = ∠FBC (proved above)
BC = BC (common side)
∠BEC = ∠CFB ( each = 900 )
By S.A.S congruence rule
Δ BEC ≅ Δ CFB
Their corresponding parts are equal
BE = CF (c.p.c.t)
4. ABC is a triangle in which altitudes BE and CF to AC and AB are equal. Show that (i) Δ ABE ≅ Δ ACF (ii) AB = AC, i.e. ABC is an isosceles triangle.
Given: BE = CF
To prove: (i) ΔABE ≅ ΔACF (ii) AB = AC
Proof: In Δ ABE and Δ ACF we have
Δ AEB = Δ AFC (each = 900 )
A = A (common angle)
BE = CF (given)
By A.A.S congruence rule
Δ ABE ≅ Δ ACF
(ii) Their corresponding parts are equal
AB = AC (c.p.c.t)
ABC is an isosceles triangle
5. ABC and DBC are two isosceles triangles on the same base BC .
Show that ΔABD ≅ ΔACD.
Solution:
Given: AB = AC and BD = CD
To prove: Δ ABD ≅ Δ ACD
Proof: In ΔABC we have
AB = AC ( ΔABE is isosceles)
∠ABC = ∠ACB
(angles opposite to equal sides are equal) —– (1)
Now in ΔBDC we have
BD = CD ( Δ BDC is isosceles)
∠CBD = ∠BCD (angles opposite to equal sides are equal) —– (2)
Adding (1) and (2) we get
Δ ABC ≅ Δ ACB
6. ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB .
Show that ΔBCD is a right triangle.
Given : AB = AC and AD = AB
To prove : ∠BCD = 900
Proof: In Δ ABC we have
AB = AC (given) ——- (1)
AD = AB (given) ——- (2)
From (1) and (2) we get
AC = AD
Now in Δ ABC we have
∠ABC + ∠ACB + ∠BAC = 1800 (angles of a Δ = 1800)
∠ACB + ∠ACB + ∠BAC = 1800
(∠ABC = ∠ACB because they are angles opposite to equal sides)
∴ 2 ∠ACB + ∠ BAC = 1800 —— (3)
Now in Δ ACD we have
∠ADC + ∠ACD + ∠CAD = 1800 (angles of a Δ= 1800)
∠ACD + ∠ACD + ∠CAD = 1800 ( ∠ADC = ∠ACD because they are angles opposite to equal sides)
∴ 2 ∠ACD + ∠CAD = 1800 —— (4)
Adding (3) and (4) we get
2 ∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 1800 + 1800
2 ( ∠ACB + ∠ACD) + ∠BAC + ∠CAD = 3600
2 (∠ACB + ∠ACD) + 1800 = 3600 ( ∠BAC + ∠CAD = 1800 because it forms a linear pair)
2 (∠ ACB + ∠ACD) = 3600 – 1800
2 ( ∠ACB + ∠ACD) = 1800
2 ∠BCD = 1800 ( ∠ACB + ∠ACD = ∠BCD)
∠BCD = 1800 / 2
∠BCD = 900
7. ABC is a right angled triangle in which A = 900 and AB = AC. Find ∠B and ∠C.
Given: ∠A = 90° and AB = AC
To prove : ∠B = ? ∠C = ?
Proof: AB = AC (given)
∠ACB = ∠ABC (angles opposite to equal sides are equal)
We know that in Δ ABC
∠A + ∠B + ∠C = 1800 ( angles of a Δ = 1800)
900 +∠ B + ∠C = 1800
∠B + ∠C = 1800 – 900
∠B + ∠C = 900
Since AB = AC we get ∠B = ∠C
∴ ∠B + ∠B = 900
2 ∠B = 900
∠B = 450
And ∠C + 450 = 900
∠C = 450
∠B = 450 and ∠C = 450
8. Show that the angles of an equilateral triangle are 600 each.
Given: AB = CA = BC
To prove : ∠A = ∠B = ∠C = 600
Proof: Since AB = BC we get
∠C = ∠A (angles opposite to equal sides are equal) ——(1)
And since AC = BC we get
∠B = ∠A (angles opposite to equal sides are equal) ——(2)
From (1) and (2) we get
∠A = ∠B = ∠C
Let ∠A = ∠B = ∠C = ∠x
∠A + ∠B + ∠C = 1800 (angles of a Δ = 1800)
∠x + ∠x + ∠x = 1800
3∠x = 1800
∠x = 1800 3 = 600
∠A = ∠B = ∠C = 600
Thus the angles of an equilateral triangle are 600 each.