**Hannah asked:**

I am doing Harold Jacobs Geometry, I am on chapter 11 lesson 3. I got to problems 39-41 and got stuck.

**Dr Callahan Answered:**

Let us start at #38.

We can see that <ABC = 120 so <ADB = 30. Therefore we know AB = BD based on Theorem 10 page 159.

**#39. **

AB = x because given

BD = x by substitution since AB = BD

<BDC = 30 (since a right triangle with one angle 60)

2BC = BD (Them 51 on page 442)

2BC = x by substitution

BC = x/2

**#40**

same logic using thrm 51 again — of course starting from what we know from #39

CD = BC*SQRT(3) Thrm 51 on page 442

BC = x/2 from #39

CD = x *SQRT(3) / 2

**#41**

We are to show how we get CD = 7/8 AB

AB = x

CD = x *SQRT(3) /2

We are to show CD/AB = 7/8 so we can divide what we have as AB and CD to get

CD/AB = (x *SQRT(3) /2 ) / x

CD/AB = SQRT(3) /2

Now you have to calculate SQRT(3) /2 to get about 0.866 and then do the math on 7/8 to get 0.875

They are not equal – but approximately equal.