In ∆ ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ∆ ABC is an isosceles triangle in which AB = AC.
In ∆ADC and ∆ADB,
AD = AD (Common)
∠ADC = ∠ADB (Each 90º)
CD = BD (AD is the perpendicular bisector of BC)
∠∆ADC ∠∆ADB (By SAS congruence rule)
∴ AB = AC (By CPCT)
∆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.
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.