Math Is Fun Forum

Mystogan

Member

Registered: 2015-08-12

Posts: 12

Circles

It would be appreciated if you could provide a detailed solution for the following question.

In a cyclic quadrilateral ABCD side AB and DC are produced to meet at point P and AD and BC are produced to meet at point Q (both P and Q are outside the circle). Find the value of ∠PTQ where PT and QT are the bisectors of ∠APD and ∠AQB respectively.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Circles

hi Mystogan

I'm looking at this question now. 

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Circles

hi Mystogan,

Sorry about the delay in replying.  We had visitors so I had to log out.

Call the four angles of the cyclic quad. a, b, c and d, AQB = 2y and APD = 2x.

The opposite angles of the quad add to 180 so a + c = b + d = 180.

If QT cuts DP at point R, then DRQ = TRP = 180 - b - y and so in triangle RTP,

PTQ = 180 - (x+180-y-b) = y - x + b ...................(1)

In triangle ABQ, 2y + a + b = 180 and
in triangle APD, a + (180-b) + 2x = 180

subtracting

2y + a + b - a - 180 + b - 2x = 0  =>  2(y-x) + 2b = 180

half this => y - x + b = 90.  using (1) this gives PTQ = 90.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Mystogan

Member

Registered: 2015-08-12

Posts: 12

Re: Circles

Thanks for making this question appear so easy, Bob!