Thread: Circle Geometry Problem

1. Circle Geometry Problem

X and Y are points on the sides BC and AC of a triangle ABC respectively such that angle AXC = angle BYC and BX = XY. Prove that AX bisects the angle BAC.

I believe the first step is to prove ABXY is a cyclic quadrilateral. Unfortunately, I dont know how!!!

2. Originally Posted by xwrathbringerx
X and Y are points on the sides BC and AC of a triangle ABC respectively such that angle AXC = angle BYC and BX = XY. Prove that AX bisects the angle BAC.

I believe the first step is to prove ABXY is a cyclic quadrilateral. Unfortunately, I dont know how!!!
I found many ways to solve and all required a good figure to show the many angles and triangles involved. Since I cannot show the figure in computers, I chose the least complicated solution for less confusion here.

>>>In any two triangles, if two of their 3 interior angles are equal in measures correspondingly, then the two triangles are similar. Or, if their 3 interior angles are equal in measures each to each, then the two triangles are similar. And, similar triangles are proportional.
>>>In an isosceles triangle, the two base angles are equal in measures. These base angles are opposite the two equal sides.
>>>A straight angle measures 180 degrees.
>>>The 3 interior angles of any triangle sum up to 180 degrees.
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Draw the figure on paper or anywhere.
Let's call the many angles involved as:

In angle A, clockwise, <1 and <2. Or, in <BAC....<XAC = <1; <XAB = <2.

In <B, clockwise, <3 and <4......Or, <YBC = <4.

In straight angle BXC, counterclockwise, <5, <6, and <7......Or, <YXC = <7.

In straight angle AYC, clockwise, <8, <9, and <10....Or, <XYC = <10.

Call the intersection of AX and BY as point O.
At point O, <BOX = <AOY = <11
and, <AOB = <YOX = <12

Finally, in <C....<BCA = <13.
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Given: BX = XY, so, triangle BXY is isosceles. Hence, <4 = <9 = say, angle alpha.

Given: <AXC = <BYC = say, angle beta. Hence, (<6 +<7) = beta = (<9 +<10).

In triangle BYC......<4 +Beta +<13 = 180deg......alpha +beta +<13 = 180deg.
IN trianle AXC....<1 +beta + <13 = 180deg also.
So, <1 = alpha.
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In triangle BOX and triangle AOY:

In straight angle BXC..... <5 = 180 - beta.
In straight angle AYC.....< 8 = 180 -beta, also.

<4 = <1 = alpha

<11 = <11

So, trangle BOX is similar to triangle AOY...and hence, proportional.
By proportion:<11 / BX = <11 / AY..... therefore, BX = AY.

Since it is given that BX = XY. then, XY = AY.....so, triangle AYX is isosceles.
therefore, <1 = < 6 = alpha.

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In triangle XOY:.....<12 = (180 -<6 -<9) = (180 -alpha -alpha) = 180 -2(alpha).

In triangle AOB..... <12 +<2 + < 3 = 180
So, (180 -2(alpha)) +(<2 + < 3) = 180
(<2 + < 3) = 2(alpha) -------------------------(1)

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Back to the similar triangles BOX and AOY:

By proportion again, <5 /BO = < 8 /AO.
since <5 = < 8, then, BO = AO......hence triangle BOA is isosceles.
So, < 3 = <2. ----------------(2)

So, tie up Eq.1 and Eq.2, ........<3 = <2 = alpha

Therefore in angle BAC, since < 1 = < 2 = alpha, then, AX bisects angle BAC ...................proven.

3. Quadrilateral ABXY is cyclic because
< AXC = < BYC

As the chords BX = XY , so arc BX = arc XY.

Thus < BAX = < CAX

So AX is the angle bisector of < BAC.