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4.8 I) Very Hard Circle Theorem Questions
4.8 I) Very Hard Circle Theorem Questions
We are going to be looking at a complex circle theorem question in this section. This question looks very hard but as soon as you understand how to answer questions like this, they become much more straightforward. Therefore, it is worth getting the working for this question down on a revision card because if a similar question comes up in the exam, you will know how to answer the question.
Example 1
A, B, R and P are four points on a circle with centre O.
A, O, R and C are four points on a different circle.
The two circles intersect at the points A and R.
CPA, AOB and BRC are straight lines.
Prove that:
A, O, R and C are four points on a different circle.
The two circles intersect at the points A and R.
CPA, AOB and BRC are straight lines.
Prove that:
angle CAB = angle ABC.
When you are answering these types of questions, you can add lines, notes and unknowns on to the diagram. The question is asking us to prove that angel CAB is equal to angle ABC. I am going to label CAB as x, and we will later be proving that ABC is also x.
I am now going to add in a line on the diagram that goes from R to O.
There is now a cyclic quadrilateral in the circle on the left (a cyclic quadrilateral is where all 4 points of the quadrilateral touch the circumference/ outside of the circle). The cyclic quadrilateral for the circle on the left is AORC. The circle theorem to do with cyclic quadrilaterals is that opposite angles in cyclic quadrilaterals add up to 180°. This means that angle CAO and angle ORC will add up to 180. And, angle AOR and RCA will also add up to 180°. The pair of opposite angles that is useful for us, is that CAO and ORC will add up to 180°. An equation for this is shown below:
Angle CAO is x, so we can replace CAO with x.
We want to find the size of angle ORC, and we are able to do this by taking x from both sides of the equation.
This tells us that angle ORC is 180° – x. I have added this information onto the diagram below.
We were told at the start of the question that CRB is a straight line. We know that angles on a straight line add up to 180°. This means that angle ORC and angle ORB will add to 180°. This gives us the equation below:
We have found that angle ORC is 180 – x, so we can replace ORC with 180 – x.
The 180 on both sides of the equation cancel each other out, which leaves us with:
The final step in finding the value of ORB is to add x to both sides of the equation.
This tells us that angle ORB is x. I have added this information to the diagram.
We were told at the start of the question that O is the centre of the circle on the right. This means that the lines OR and OB are radii for the circle on the right. Therefore, the triangle OBR is an isosceles triangle as two of the sides of the triangle are the same length; OR and OB are the same length. The base angles for isosceles triangles are the same size and this means that angle ORB and angle OBR are the same size. We have found that angle ORB is x, which means that angle OBR is also x. I have added this onto the diagram below.
From looking at the above diagram, we can see that angles CAB and ABC are both x. Therefore, we have proved that angle CAB and ABC are equal to one another.
It may take a few attempts to fully get all of the steps for this question. Therefore, it definitely worth going through this example a few times with a break in between each time.
It may take a few attempts to fully get all of the steps for this question. Therefore, it definitely worth going through this example a few times with a break in between each time.