1. In the figure to the left, D and E are mid-points of AB and AC respectively. Prove that ABC ~ ADE.
Solution: We will use the SAS method of proving similarity in this case:
angle A = angle A (common angle)
AD/AB = AE/AC = 1/2 (D and E are mid-points of AB and AC)
=> ABC ~ ADE
2. In the figure below (left), BAC is a right angled triangle with a right angle at A. AD is the perpendicular from vertex A to side BC.
Prove that ABD ~ CAD.
Solution: In right triangle BAC, angle B = 90o - angle C ... (1)
In right triangle CDA, angle CAD = 90o - angle C ... (2)
By (1) and (2),
angle B = angle CAD
Also, angle BDA = angle CDA (both 90o)
Thus, by AA similarity condition, ABD ~ CAD.