The diagram shows a point X on a circle and XY is a tangent to the circle. Points A, B and C lie on the circle such that XA bisects angle YXB and YAC is a straight line. The lines YC and XB intersect at D.
(i) Prove that AX = AB
(ii) Prove the CD bisects angle XCB.
(iii) Prove that the triangles CDX and CBA are similar.
Word to the wise: In questions like these, I normally find it useful to redraw the diagram in the answer sheet. That way, I can annotate on the diagram as part of the answer. But more importantly, when I have completed drawing it out as per the description, I have a better understanding of the diagram. Hence, I would strongly recommend that you do draw it out, even if it seems easy. Your understanding deepens a lot more when you do that.
Okay, now on to the answers…
(i) To show that AX = AB, we will need to prove that triangle XAB is isosceles.
Let angle YXA = x°
Angle YXA = Angle XBA (alternate segment theorem)
Hence, angle AXB = angle XBA
Therefore, triangle XAB is isosceles and AX = AB.
(ii) Angle YXB = Angle XCB (alt seg)
Therefore, angle XCB = 2x°
Since we have shown that angle AXB = angle XBA = x°
And that angle XCA = angle XBA (angles in the same segment) = x°
Therefore, angle ACB = angle XCA = x°
Hence, the line CD bisects the angle XCB.
(iii) Since angle CXD = angle CAB (A)
angle XCD = angle ACB (A)
Therefore, angle CDX = angle CBA (A)
Hence, triangle CDX is similar to CBA (AAA)
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