Question

the bisector of angle B of an isosceles triangle ABC with AB=AC meets the circumcircle of triangle ABC at P if AP AND BC produced meet at Q
prove that CQ=CA

Answer 1

triangle ABC is isosceles, with angle B = angle C
Call the bisected angle at B = x (so angle B and angle C are both 2x)

use these theorems:
angle at Q formed by 2 secants = (angle of outer arc - angle of inner arc)/2
in other words angle Q= ( arc AB - arc PC)/2

also, inscribed angle PBC = (arc PC)/2. that is, x= arc PC /2
similarly 2x = arc AB/2

Answer 2

you can now find angle Q in terms of x

the angle QCA in triangle AQC is 180-2x
you know Q, you know 180-2x so you can find the 3rd angle, angle CAQ
you will find angle CAQ = angle AQC, so the triangle AQC is isosceles, with sides
CQ= CA

can you fill in the details?

Answer 3

thank u so much @phi

Answer 4

Here are the theorems, in case you have not seen them
http://www.regentsprep.org/Regents/math/geometry/GP15/CircleAngles.htm

the secant one used here is at the bottom of the page.

Answer 5

k

Answer 6

also i don't understand this part,

you can now find angle Q in terms of x

the angle QCA in triangle AQC is 180-2x
you know Q, you know 180-2x so you can find the 3rd angle, angle CAQ
you will find angle CAQ = angle AQC, so the triangle AQC is isosceles, with sides
CQ= CA

can you fill in the details?

Answer 7

Given
isosceles triangle ABC with AB=AC

that means angle ABC = angle ACB. I labeled the measure of these angles = 2x
(that makes the measure of the bisected angle PBC = x )

Answer 8

angle ACB is an "inscribed angle". It is equal to 1/2 the arc AB
ACB= 2x, so we have: arc AB *1/2 = 2x and arc AB= 4x

arc AP= 2x by the same reasoning

angle Q= ( arc AB - arc PC)/2
angle Q= ( 4x - 2x)/2 = 2x/2 = x

the angles in triangle AQC are : 180-2x, x, and a 3rd angle Y
Y+x+180-2x = 180
Y= 180-180+x = x

we see angle Y (that is, angle CAQ) has measure x.
also, angle Q (that is, angle AQC) has measure x
that means that triangle AQC is isosceles, with legs CA= CQ