DP=CP=5

Triangle DAP is similar to triangle CPQ, and CQ=2.5 and BQ=7.5 are obtained.

In the triangle BAQ, we can get AB= 10, BQ=7.5, AQ= 12.5.

In the triangle CBQ, BC= 10, CP=5, BP=5 times the root number 5 can be obtained.

In the triangle CPQ, CP=5, CQ=2.5, PQ=(5 times the root number 5)/2 are obtained.

In the triangle ADP, we get AD= 10, DP=5, AP=5 times the root number 5.

In right triangle APQ and right triangle PCQ, PQ/AP=CQ/CP= 1/2, and ∠CPQ=∠PAQ is obtained.

Using the angle relation, we get ∠DAP=∠CPQ=∠CBP=∠PAQ.

Because the diagonals are equal, ∠BTQ=∠ATP, the triangle BTQ is similar to the triangle ATP.

Using the similar triangle relation, we get:

At/Pt = Bt/Qt = AB/PQ =10/[(5th radical number 5)/2]=4/ radical number 5.

Get AT=4PT/ radical number 5, QT= radical number 5*BT/4.

And because AT+QT=AQ= 12.5, BT+PT=BP=5 times the root number 5, the equation holds:

4PT/ radical number 5+ radical number 5*BT/4= 12.5

BT+PT=BP=5 times the root number 5.

BT=20 root number 5/ 1 1.

PT=25 times the root number 5/ 1 1

So BT/PT=6/5.