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)
(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.
