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How
to Write a Solution
Follow the Lemmas
by
Richard Rusczyk
Often you will have to prove multiple preliminary items before tackling the main problem. In writing a proof, we often choose to separate these parts from the main proof by labeling each as a 'Lemma' and clearly delimiting the lemma and its proof from the rest of the solution.
Here's a sample problem with two different solutions that employ lemmas. We've used a little overkill in writing the solutions with lemmas to highlight how well we can clarify solutions with lemmas. Both of these solutions are made significantly easier to read by clearly breaking the solution into pieces.
Here are two solutions:
How Not to Write the Solution 1: Let the line through A parallel to BC meet line BM at J. Let the line through J parallel to AB meet line BC at K. Let MN hit AB at X and AC at Y.
Since JK || AB and AJ || BK, JKBA is a parallelogram. Since <ABM is half the exterior angle ABK, we have
<ABM = (<A + <C)/2.
From right triangle BAM, we find
<BAM = 90 - <ABM = <B/2.
Since AJ || BC, we have <JAB = <B, so MA bisects <BAJ. Thus, <JAM = <BAM and triangles BAM and JAM are congruent by ASA. Thus, AJ = AB and parallelogram JKBA is a rhombus.
The diagonals of a rhombus bisect each other, so triangles AMJ and KMB are congruent. Thus, the altitudes from M to AJ and BK are equal and M is equidistant from lines AJ and line BK. By symmetry, this is also true for N. Thus, MN || BC and MN is equidistant from AJ and BC.
Since MX || BK, triangle AMX ~ triangle AKB. Since AM = AK/2, we have MX = KB/2. Since JKBA is a rhombus, KB = AB, so MX = AB/2, as desired.
By symmetry, we also have NY = AC/2 from. Since XY is the midline of ABC parallel to side BC, we have XY = BC/2. Thus,
MN = MX + XY + YN = AB/2 + AC/2 + BC/2,
as desired.
Here's the same solution, written with lemmas. Notice how the lemmas break up the solution into pieces nicely:
How to Write the Solution 1: Let the line through A parallel to BC meet line BM at J. Let the line through J parallel to AB meet line BC at K. Let MN hit AB at X and AC at Y.
We will show that MX = AB/2, XY = BC/ 2, and YN = AC/2, from which the desired result follows.
Lemma
1: JKBA is a rhombus.
Proof: Since JK || AB and AJ || BK, JKBA is a parallelogram.
Hence, we need only prove that a pair of consecutive sides are equal
to conclude JKBA is a rhombus.
Since <ABM is half the exterior angle ABK, we have
<ABM = (<A + <C)/2.
From right triangle BAM, we find
<BAM = 90 - <ABM = <B/2.
Since AJ || BC, we have <JAB = <B, so MA bisects <BAJ. Thus, <JAM = <BAM and triangles BAM and JAM are congruent by ASA. Thus, AJ = AB and parallelogram JKBA is a rhombus.
---------------end lemma---------------
Lemma
2: MN || BC and MN is equidistant from lines AJ and BC.
Proof: The diagonals of a rhombus bisect each other,
so triangles AMJ and KMB are congruent. Thus, the altitudes from M to
AJ and BK are equal and M is equidistant from lines AJ and line BK.
By symmetry, this is also true for N. Thus, MN || BC and MN is equidistant
from AJ and BC.
---------------end lemma---------------
Lemma 3
: MX = AB/2.
Since MX || BK, triangle AMX ~ triangle AKB. Since AM = AK/2, we have
MX = KB/2. Since JKBA is a rhombus (Lemma 1), KB = AB, so MX = AB/2,
as desired.
---------------end lemma---------------
By symmetry, we also have NY = AC/2 from Lemma 3. Since XY is the midline of ABC parallel to side BC, we have XY = BC/2. Thus,
MN = MX + XY + YN = AB/2 + AC/2 + BC/2,
as desired.
Here is a second solution:
How Not to Write the Solution 2: The external angle bisectors of B and C meet at E, an excenter of triangle ABC opposite vertex A as shown. Let P and Q be the feet of the perpendiculars from E to lines AC and AB and let X be the point of tangency of the excircle centered at E to segment BC.
Since E is on the angle bisector of <BAC, we have from right triangle EPA:
<AEP = 90° - <BAC/2 = (<ABC + <ACB)/2.
From triangle CEB, we have
<MEC = <CEB = 180° - <CBE - <BCE.
Since E is on the angle bisector of <CBQ, we have
<CBE = (180° - <ABC)/2 = 90° - <ABC/2.
Similarly, <BCE = 90° - <ACB/2. Substituting these in our expression for <MEC gives the desired
<MEC = (<ABC + <ACB)/2.
Thus, <AEP = <MEC, as desired.
Since
<APE = <ANE = <AQE = <AME,
points N, Q, M, and P are on the circle with AE as diameter.
Since <AEP = <MEC = <MEN, we have AP = MN since they subtend equal arcs of this circle.
Since tangents to a circle from a point are equal, we have AQ = AP, BX = BQ, and CX = CP. Thus,
AC + CX = AC + CP = AP = AQ = AB + BQ = AB + BX
From AC + CX = AB + BX, we have
CX = AB - AC + BX = AB - AC + (BC - CX), so
CX = (AB + BC - AC)/2.
Thus,
AP = AC + CP = AC + CX = (AB + BC + AC)/2.
Since MN = AP, we have
MN = AP = (AB + BC + AC)/2,
as desired.
The solution above is essentially 3 main steps, which we tie together to complete our solution. With lemmas we can clearly define the 3 parts, the proofs of each, and how they are combined to give the final result:
How to Write the Solution 2: The external angle bisectors of B and C meet at E, an excenter of triangle ABC opposite vertex A as shown. Let P and Q be the feet of the perpendiculars from E to lines AC and AB and let X be the point of tangency of the excircle centered at E to segment BC.
We will show that AP = (AB + BC + AC)/2, and that MN = AP because they are chords that subtend equal arcs of the circle circuscribed about ANPEQM. The desired result follows.
Lemma
1 :
<AEP = <MEC.
Proof: We will show that both equal (<ABC + <ACB)/2.
Since E is on the angle bisector of <BAC, we have from right triangle EPA:
<AEP = 90° - <BAC/2 = (<ABC + <ACB)/2.
From triangle CEB, we have
<MEC = <CEB = 180° - <CBE - <BCE.
Since E is on the angle bisector of <CBQ, we have
<CBE = (180° - <ABC)/2 = 90° - <ABC/2.
Similarly, <BCE = 90° - <ACB/2. Substituting these in our expression for <MEC gives the desired
<MEC = (<ABC + <ACB)/2.
Thus, <AEP = <MEC, as desired.
---------------end lemma---------------
Lemma
2:
ANPEQM is cyclic.
Proof: Since
<APE = <ANE = <AQE = <AME,
points N, Q, M, and P are on the circle with AE as diameter.
---------------end lemma---------------
Lemma 3
:
AP = (AB + BC + AC)/2.
Proof: Since tangents to a circle from a point are
equal, we have AQ = AP, BX = BQ, and CX = CP. Thus,
AC + CX = AC + CP = AP = AQ = AB + BQ = AB + BX
From AC + CX = AB + BX, we have
CX = AB - AC + BX = AB - AC + (BC - CX), so
CX = (AB + BC - AC)/2.
Thus,
AP = AC + CP = AC + CX = (AB + BC + AC)/2.
---------------end lemma---------------
Since <AEP = <MEC = <MEN from Lemma 1, we have AP = MN since they subtend equal arcs of our circumcircle of ANPEQM from Lemma 2. Combining this with Lemma 3, we have
MN = AP = (AB + BC + AC)/2,
as desired.
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