Final answer:
The perimeter of ∆ABH is 2a + 2b.
Step-by-step explanation:
Given that ∆ABC is an isosceles triangle, we can determine the perimeter of ∆ABH.
Since ∆ABC is isosceles, −BAC = −BCA. This means that angle −CAB is equal to angle −ACB.
When a triangle is isosceles, the two equal sides are opposite the two equal angles.
In ∆ABH, we have AD = a cm and HD = b cm. Since AD and HD are equal sides of the isosceles triangle, they are opposite equal angles. Therefore, angle ADH = angle HDA.
Therefore, the perimeter of ∆ABH is a + b + a + b = 2a + 2b.