Video: AQA GCSE Mathematics Higher Tier Pack 5 β€’ Paper 1 β€’ Question 21

𝐴, 𝐡, and 𝐢 are points on a circle and 𝑂 is the center of the circle. 𝐷𝐢𝐸 is a tangent to the circle. 𝐢𝐡 bisects the angle 𝐴𝐢𝐸. Angle 𝐡𝐢𝐸 = π‘₯Β° and angle 𝑂𝐡𝐢 = (90 βˆ’ π‘₯Β°). Prove that 𝐴𝐡 = 𝐡𝐢.

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Video Transcript

𝐴, 𝐡, and 𝐢 are points on a circle and 𝑂 is the center of the circle. 𝐷𝐢𝐸 is a tangent to the circle. 𝐢𝐡 bisects the angle 𝐴𝐢𝐸. Angle 𝐡𝐢𝐸 is equal to π‘₯ degrees and angle 𝑂𝐡𝐢 is equal to 90 minus π‘₯ degrees. Prove that 𝐴𝐡 is equal to 𝐡𝐢.

In order to prove that 𝐴𝐡 is equal to 𝐡𝐢, we need to show that triangle 𝐴𝐡𝐢 is isosceles. If this is the case, it will have two equal angles at 𝐡𝐴𝐢 and 𝐡𝐢𝐴. And this will mean the sides 𝐴𝐡 and 𝐡𝐢 have to be equal in length. Let’s first see if there’s any useful information that we can use in the question itself.

We’re told that 𝐢𝐡 bisects the angle 𝐴𝐢𝐸; that is to say, the line 𝐢𝐡 cuts the angle 𝐴𝐢𝐸 in half. For this to be the case, that must mean that angle 𝐡𝐢𝐸 is equal to angle 𝐡𝐢𝐴. They’re both π‘₯ degrees. Next, we need to recall any circle theorems that might help us here. Straightaway, I can see one that might be useful. We know that two radii or radiuses form an isosceles triangle. 𝑂𝐡 and 𝑂𝐢 are lines which join the center of the circle to a point on its circumference. That means they’re the radii of the circle and triangle 𝑂𝐡𝐢 is isosceles.

This means we know the 𝑂𝐢𝐡 and 𝑂𝐡𝐢 are equal. They’re both 90 minus π‘₯ degrees. We also know that angles in a triangle sum to 180 degrees. So we can use this information to calculate the size of angle 𝐡𝑂𝐢 marked. Angle 𝐡𝑂𝐢 can be found by subtracting angle 𝑂𝐡𝐢 and angle 𝑂𝐢𝐡 from 180 degrees. 180 minus 90 minus 90 is zero. Negative negative π‘₯ is simply π‘₯, minus negative π‘₯ is plus π‘₯. And we can see that angle 𝐡𝑂𝐢 is equal to two π‘₯ degrees.

We can see now that angle 𝐡𝑂𝐢 and 𝐡𝐴𝐢 are subtended from points on the circumference of the circle. We can, therefore, use the circle theorem that says the angle at the center is twice the angle at the circumference. Another way to think about this is that the angle at the circumference is half the angle at the center. Angle 𝐡𝐴𝐢 is half angle 𝐡𝑂𝐢. It’s a half of two π‘₯ which is simply π‘₯ degrees.

We have now shown that 𝐴𝐡𝐢 is an isosceles triangle with equal angles at 𝐡𝐴𝐢 and 𝐡𝐢𝐴. This means in turn that the sides 𝐴𝐡 and 𝐡𝐢 must also be of equal length. We have proven that 𝐴𝐡 is equal to 𝐡𝐢.

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