Solution to #12 on Page 234

For this problem you must remember the theorem that states that the altitudes of similar triangles are proportional. To keep he post short I have summarized the solution. The diagram to te right is what I will be referring to for this solution. Some points were in the description in the question and others are introduced as a result of the auxiliary lines.

You know that QDC is similar to QAC since their angles are equal. Thus, AQ/DQ = BQ/CQ = AB/DC. The ratio of AB/DC is 3/2 from the given. Moreover, the altitudes of similar triangles are also proportional and so JQ/KQ = AQ/DQ. Therefore, JQ/KQ = 3/2. We are first interested in finding the length of JQ, which consists of JK+KQ. JK is 1.5 inches since the distance between DC and AB is 1.5 inches. Thus, 1.5 +KQ = JQ. Then, (1.5 + KQ)/KQ = 3/2. You can now solve this equation for KQ. When you have KQ then you can know JQ, which is the distance of Q from AB.

Now consider my auxiliary line through P that is perpendicular to both AB and DC and serves as the distance between AB and DC just as JK does. Now due to alternate interior angles formed by the dagonals and equal opposite angles you know that ABP and CDP are similar triangles. Also, you will find that PY and PX are altitudes for those triangles, respectively. If the triangles are similar then the altitudes are similar. Thus, AB/DC = PY/PX. AB/DC = 3/2 and so PY/PX = 3/2. We know that PY+PX = 1.5 inches from the given. So, you have two equations and two variables. Now all you have to do is find PY.