Two geometry doubts. Please help me!?

Question

1) If the longer base of an isosceles trapezium equals a diagonal and the shorter base equals the altitude, prove that the ratio of the shorter base to the longer base is 3 : 5

2) In a triangle ABC, the in-circle touches the sides BC, CA and AB at D, E and F respectively. If the radius of the in-circle is 4 units and if BD, CE and AF are consecutive integers, find AB, BC and CA.

I'll give a hint for the second one:
Use BD = x - 1, CE = x, AF = x + 1
Area of triangle = (1/2) * (In-radius) * (Perimeter)

Answer 1

1) Let "a" is the length of the longer base (and the diagonal), "b" - of the shorter (and height). Draw a height from one of the upper vertexes and take the right triangle with legs: the height (b) and the distance from its foot to the farther vertex of the longer base (b + (a-b)/2 = (a+b)/2) and hypotenuse a:
((a+b)/2)^2 + b^2 = a^2, or (a + b)^2/4 = a^2 - b^2, or
(a + b)^2 = 4(a + b)(a - b), or a + b = 4(a - b), or
b + 4b = 4a - a, or 3a = 5b, or b/a = 3/5.

2) Easily follows that the sides are also consecutive integers (c-1, c, c+1). Here the middle side "c" satisfies, according your hint and Heron's formula for the area (the area expressed twice):
36c^2 = (3c^2/4)(c^2/4 - 1), i.e. c^2/4 - 1 = 48, and
c^2 = 49*4 = 14^2, so c = 14. The sides are 13, 14, 15.

Answer 2

how do i solve if asker dont give hints(on Q2)?