Page 3 - ln1

Solution

(i) It is given that in triangle ABC, AB = AC
∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are equal)
1 1
∠ACB = ∠ABC
2 2
∠OCB = ∠OBC
∴ OB = OC (Sides opposite to equal angles of a triangle are also equal)
(ii) In ΔOAB and ΔOAC,
AO =AO (Common)
AB = AC (Given)
OB = OC (Proved above)
Therefore, ΔOAB ≅ ΔOAC (By SSS congruence rule)
∠BAO = ∠CAO (CPCT)
∴ AO bisects ∠A.

2.ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal
sides AC and AB respectively (see the given figure). Show that these altitudes are
equal.

Solution
In ΔAEB and ΔAFC,
∠AEB and ∠AFC (Each 90º)
∠A = ∠A (Common angle)
AB = AC (Given)
∴ ΔAEB ≅ ΔAFC (By AAS congruence rule)
∴ BE = CF (By CPCT)

3.ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such
that AD = AB (see the given figure). Show that ∠BCD is a right angle.