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
(i) It is given that in triangle ABC, AB = AC
∴ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are equal)
∴\(\frac{1}{2}\) ∠ACB= \(\frac{1}{2}\) ∠ABC
∴ ∠OCB =∠OBC
∴ OB = OC (Sides opposite to equal angles of a triangle are also equal)
(ii) In ∆OAB and ∆OAC,
AO =AO (Common)
AB = AC (Given)
OB = OC (Proved above)
Therefore, ∆OAB ∆OAC (By SSS congruence rule)
∠BAO = ∠CAO (CPCT)
∴ AO bisects A.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.
ABC and DBC are two isosceles triangles on the same base BC (see Fig). Show that ∠ABD = ∠ACD.
In ∆ ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ∆ ABC is an isosceles triangle in which AB = AC.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.