Geometry Worksheet 3 – Proofs
1. 𝐴𝐵 and 𝐴𝐶 are chords of a circle, and 𝐴𝐷 is a line that bisects angle ∠𝐵𝐴𝐶, where 𝐷 is a point
on the circumference of the circle. Prove that 𝐵𝐷 = 𝐶𝐷. (Hint: perhaps draw the tangent at
point 𝐷)
2. [Source: UKMT Mentoring] Let 𝐴𝐵𝐶𝐷 be a square. Let 𝑀 be the midpoint of 𝐴𝐵 and 𝐾the point
on 𝐴𝐷 such that ∠𝐾𝑀𝐶 is a right angle. Prove that the sides of triangle 𝐾𝐷𝐶 are in the ratio
3: 4: 5 (Hint: there might be some similar triangles in there!).
3. [Source: UKMT Mentoring] A square 𝐴𝐵𝐶𝐷 of length 1 is inscribed inside a circle. A point 𝑃 is on
the arc 𝐶𝐷. Use Ptolemy’s Theorem to show that 𝑃𝐴 × (𝑃𝐴 + 𝑃𝐶) = 𝑃𝐵 × (𝑃𝐵 + 𝑃𝐷).
4. [Source: UKMT Mentoring] Circle 𝛾1 lies inside circle 𝛾2 and touches it at 𝐴. From a point 𝑃
(distinct from 𝐴) on 𝛾2 , chords 𝑃𝑄 and 𝑃𝑅 of 𝛾2 are drawn touching 𝛾1 at 𝑋 and 𝑌 respectively.
Show that ∠𝑄𝐴𝑅 = 2∠𝑋𝐴𝑌.
5.
[Source: BMO Round 1] In the cyclic quadrilateral 𝐴𝐵𝐶𝐷, the diagonal 𝐴𝐶 bisects the angle 𝐷𝐴𝐵.
The side 𝐴𝐷 is extended beyond 𝐷 to a point 𝐸. Show that 𝐶𝐸 = 𝐶𝐴 if and only if 𝐷𝐸 = 𝐴𝐵
[Note: ‘if and only if’ means you have to proof both ways, i.e. that if 𝐶𝐸 = 𝐶𝐴 then 𝐷𝐸 = 𝐴𝐵 and
if 𝐷𝐸 = 𝐴𝐵 then 𝐶𝐸 = 𝐶𝐴]
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Geometry Worksheet 3 - ANSWERS
1.
Let ∠𝐵𝐴𝐶 = ∠𝐶𝐴𝐷 = 𝑎. Suppose we draw the tangent at point 𝐷. Then by the Alternate Segment
Theorem, the angle between the tangent and each of the chords 𝐵𝐷 and 𝐶𝐷 is 𝑎. Since these angles
are the same, by an argument of symmetry, it must be that 𝐵𝐷 = 𝐶𝐷.
2.
We have the following diagram on the right. Without loss of generality, let the
side length of the square be 2.
If ∠𝑀𝐶𝐵 = 𝑎, then ∠𝐶𝑀𝐵 = 90° − 𝑎 and so ∠𝐴𝑀𝐾 = 𝑎. Thus triangles 𝐴𝐾𝑀
and 𝑀𝐵𝐶 are similar. This allows us to determine that 𝑀𝐶 = 2𝐾𝑀.
Using Pythagoras, 𝐵𝐶 2 + 𝑀𝐵2 = 𝑀𝐶 2 , so 𝑀𝐶 = √5. And since 𝑀𝐶 = 2𝐾𝑀,
𝐾𝑀 =
√5
2
5
. We can again use Pythagoras on triangle 𝐾𝑀𝐶 to find that 𝐾𝐶 = .
3
2
3 4 5
2
2 2 2
Finally using Pythagoras in triangle 𝐾𝐷𝐶, 𝐾𝐶 = . The lengths of 𝐾𝐷𝐶 are , ,
which are in the ratio 3: 4: 5.
3.
We know that 𝐴𝐶 = 𝐵𝐷 = √2. Then applying Ptolemy’s Theorem to cyclic quadrilateral 𝐴𝐵𝐶𝑃, we
have √2 𝑃𝐵 = (1 × 𝑃𝐴) + (1 × 𝑃𝐶) = 𝑃𝐴 + 𝑃𝐶. Similarly using quadrilateral 𝐴𝐵𝑃𝐷, we have
√2 𝑃𝐴 = 𝑃𝐵 + 𝑃𝐷. Dividing one equation by the other and then cross-multiplying, we’re done.
4.
Let ∠𝑋𝐴𝑌 = 𝑎. Then by the Alternate Segment Theorem, ∠𝑃𝑋𝑌 = ∠𝑃𝑌𝑋 = 𝑎. Thus ∠𝑃𝑋𝑌 = 180 −
2𝑎 (internal angles of a triangle). But 𝑃𝑄𝐴𝑅 is a cyclic quadrilateral, so ∠𝑄𝐴𝑅 = 180 − (180 − 2𝑎) =
2𝑎. Thus ∠𝑄𝐴𝑅 = 2∠𝑋𝐴𝑌.
5.
[Note that, like Q2 and Q4, I haven’t seen the ‘model solution’ for this. The model solutions
may have a more elegant way of approaching this question, although in some ways I think
it’s more beneficial for you to see my ‘internal dialogue’ as I solved this question in a fairly
methodical away, rather than some elegant trick for which your reaction might be ‘I would
have never thought of doing that.’]
I started off by putting the information
given into a diagram:
My immediate reaction was that I had
two chords on the same tangent, so
could apply the Alternate Segment
Theorem twice. This gives ∠𝐷𝐶𝐸 =
∠𝐵𝐶𝑋 = 𝑎. Since these two angles are
the same, then by symmetry, it must
be the case that 𝐶𝐷 = 𝐵𝐶. I let
∠𝐷𝐶𝐴 = 𝑏, because we’d then have
another chord in contact with the
tangent (i.e. 𝐶𝐴) and could apply the
Alternate Segment Theorem again.
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Thus ∠𝐴𝐵𝐶 = 𝑎 + 𝑏. We also find that ∠𝐶𝐷𝐸 = 𝑎 + 𝑏 because it’s the external angle of
triangle 𝐴𝐶𝐷. All this information I gleaned is displayed below:
This already looks quite promising, because there’s the distinct whiff of similar and
congruent triangles. I can see that 𝐴𝐶𝐸 and 𝐴𝐷𝐸 are similar (the tell-tale sign is one triangle
embedded within another with ∠𝐶𝐴𝐷 = ∠𝐷𝐶𝐸), which I’m hoping will help.
We now have two proofs required:
a. Forwards proof: Show that if 𝑪𝑬 = 𝑪𝑨 then 𝑫𝑬 = 𝑨𝑩.
If 𝐶𝐸 = 𝐶𝐴 then 𝐶𝐸𝐴 is an isosceles triangle and ∠𝐷𝐸𝐶 = 𝑎.We can see that triangles 𝐷𝐸𝐶
and 𝐴𝐵𝐶 are not only similar but congruent (I looked at these triangles given that they
involve the sides we want to prove something about), because 𝐶𝐸 = 𝐶𝐴 and 𝐷𝐶 = 𝐵𝐶 and
∠𝐵𝐴𝐶 = ∠𝐷𝐸𝐶 and ∠𝐴𝐵𝐶 = ∠𝐸𝐷𝐶. Thus we find that 𝐷𝐸 = 𝐴𝐵.
b. Backwards proof: Show that if 𝑫𝑬 = 𝑨𝑩 then 𝑪𝑬 = 𝑪𝑨.
𝐷𝐸
𝐶𝐸
𝐷𝐸
Triangles 𝐴𝐶𝐸 and 𝐷𝐸𝐶 are similar, so 𝐷𝐶 = 𝐴𝐶. If we can show that 𝐷𝐸 = 𝐷𝐶, then 𝐷𝐶 =
𝐶𝐸
1 = 𝐴𝐶 , and thus we’d have shown 𝐶𝐸 = 𝐶𝐴 as required. We’d have 𝐷𝐸 = 𝐷𝐶 if ∠𝐷𝐸𝐶 = 𝑎,
thus the proof boils down to showing this is the case. This is easy to show by congruent
triangles again. Since 𝐷𝐸 = 𝐴𝐵 and since ∠𝐸𝐷𝐶 = ∠𝐴𝐵𝐶, triangle 𝐴𝐵𝐶 is congruent to
𝐸𝐷𝐶. Thus ∠𝐷𝐸𝐶 = ∠𝐵𝐴𝐶 = 𝑎.
𝐷𝐸
𝐶𝐸
In hindsight, I could have also used the fact that 𝐷𝐶 = 𝐴𝐶 in the forward proof as well, since if
𝐶𝐸 = 𝐴𝐶, then 𝐷𝐸 = 𝐷𝐶. It would have then been easy using congruent triangles to show
that 𝐷𝐸 = 𝐴𝐵.
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