Thread: Ptolemy's Theorem and Cyclic Quadrilaterals

In need of some help for 3 problems that I can't seem to understand. I would greatly appreciate some help.

Point P on side AB of a right Triangle, ABC, is placed so that BP = PA = 2. Point Q is on hypotenuse AC such that PQ is perpendicular to AC. If CB = 3 what is the measure of BQ?

What is the area of quadrilateral CBPQ?

As P is translated from B to A along BA, find the range of values of BQ, where PQ remains perpendicular to CA.


I don't understand the part where you translate P along BA at all. I can't seem to even draw the diagram correctly for the quadrilateral. I am also having a hard time drawing the diagram to even solve the first part, I just know the pyhgreoen theorem might help out. Thank you so much for the help.

i think the diagram comes likes this

Oh thank you so much for the diagram ! This helps a lot but I'm still wondering how I can find the measure of BQ when I connect Point B to Q. When I do that is the angle PBQ a right angle ?

i doubt it.

i think you were able to find the length of AQ.
If so, since you know AB and you can use cosine equation to find BQ

Originally Posted by zzz123

Oh thank you so much for the diagram ! This helps a lot but I'm still wondering how I can find the measure of BQ when I connect Point B to Q. When I do that is the angle PBQ a right angle ?

1. You are dealing with similar right triangles:



and



2. The distance BQ is the hypotenuse of the right triangle

I'm not sure I follow. How do I find length of AQ ? Im given AB but unsure of what Angle PAQ.

Oh seeing that they are similar helped a lot ! But now im unsure on how to find the length of QF and FC.

Thanks again for both your help.