Some Length Conditions on a Triangle

Posted on by Lasitha Jayasinghe

In this post we will have a look at three really common triangle length conditions. They look almost the same but their slightly different definitions invoke highly contrasting configurations. After introducing each condition, and looking at how it can be interpreted, we solve (at least partially) some contest problems involving those conditions.

1. Equal Lengths on non-base sides – DB = EC.

This condition is really common and it is actually quite simple to handle.

Theorem 1

Let ABC be a triangle, with M the major arc midpoint of BC in (ABC). Let D and E be points on sides AB and AC respectively such that DB=EC. Then points A , M , D and E are concyclic.

Proof : Consider the two triangles \triangle MDB and \triangle MEC. Since M is the major arc midpoint of BC, it is clear that MB = MC and \angle DBM = \angle ABM = \angle ACM = \angle ECM. Now, due to the provided length condition, DB=EC which is sufficient to imply that \triangle MDB \cong \triangle MEC.

Then, \angle ADM = 180 - \angle MDB = 180 - \angle MEC = \angle AEM which implies the desired claim.

One can further note that in fact, M should also be the major arc midpoint of DE in (ADE) since due to the established congruency we have MD=ME. This in fact implies that M is the center of spiral similarity mapping DE \to BC.

Example 2 (GOTEEM 2020/5)

Let ABC be a triangle and let B_1 and C_1 be variable points on sides \overline{BA} and \overline{CA}, respectively, such that BB_1 = CC_1. Let B_2 \neq B_1 denote the point on (ACB_1) such that BC_1 is parallel to B_1B_2, and let C_2 \neq C_1 denote the point on (ABC_1) such that CB_1 is parallel to C_1C_2. Prove that as B_1, C_1 vary, the circumcircle of \triangle AB_2C_2 passes through a fixed point, other than A.

To one who hasn’t seen this before, the condition and also the conclusion could appear quite overwhelming. But now that we know the major arc midpoint M of BC is inherently connected with this configuration, we throw in this point. After this addition, we naturally stumble upon the following question.

Is M the desired fixed point?

We work towards a solution keeping this intuitive nudge in mind. Now, we know before hand that points A , B_1 , C_1 and M are concyclic. To proceed from here, we first need to show the following key claim.

Claim : Quadrilateral B_1C_1B_2C_2 is cyclic (with circumcenter M).

Proof : Let R = (ABC_1) \cap (AB_1C). Then, BB_1=CC_1 , \measuredangle RB_1B = \measuredangle RCA and \measuredangle RBA = \measuredangle CC_1R. Thus, \triangle RB_1B \cong \triangle RCC_1. Now, this means there exists a spiral similarity centered at R mapping BB_1 \mapsto CC_1. Thus, there must also exist a spiral similarity centered at R mapping BC_1 \mapsto CB_1. This allows us to conclude that \triangle RBC_1 \sim \triangle RCB_1. Let C_2 ' = \overline{RB_1} \cap (AC_1B) and B_2' = \overline{RC_1} \cap (AB_1C). Then,

\measuredangle CB_1R = \measuredangle C_1BR = \measuredangle C_1C_2'R

and thus, \overline{C_1C_2'} \parallel \overline{CB_1} which implies that C_2'=C_2. Similarly we conclude that B_2'=B_2. Thus, lines \overline{B_1C_2} and \overline{B_2C_1} intersect at R. Clearly, R is the arc midpoint of minor arc BC_1 (since RB=RC_1 due to the previously establish congruency). So, \overline{C_2B_1} is the \angle C_1C_2B-bisector. Now, we can note that,

2\measuredangle C_1C_2B_1 = \measuredangle C_1CB = \measuredangle C_1AB_1 = \measuredangle C_1MB_1

so it follows that C_2 lies on \omega. A similar argument implies that B_2 also lies on \omega which finishes the proof of the claim.

The finish from here is a simple angle chase to show that M lies on (AB_2C_2) which is (gladly) left as an exercise to the reader.

One can also attempt the problems ; [EGMO 2014/2] and the following nice problem by TestX01. These problems look at a further specific case of this setup where BD=BC=CE.

2. Equal segments to base dissection – BE=BD and CD=CE.

Here, again the way to interpret the given length condition is a concyclicity, which looks eerily similar to the previous one.

Theorem 3

Let ABC be a triangle with point D on side BC and incenter I. Let E and F be points on sides AB and AC such that BD=BE and CD=CE. Then points A , E , F and I are concyclic, with I in fact being the center of (DEF). Also, points \overline{BI} \cap \overline{DF} and \overline{CI} \cap \overline{DE} lie on (AEF).

Proof : We again start with some congruence. In triangles, BID and BIE, IB is a common side with \angle EBI = \angle IBD. Since due to our length condition, BD=BE which implies that \triangle BID \cong \triangle BIE. Thus, ID=IE. A similar argument also show that ID=IF so I is indeed the circumcenter of \triangle DEF.

Now, note that \angle EIF = 2\angle EDF = 2(180 - \angle BDE - \angle CDF) = 180 - \angle FAE, so I indeed lies on (AEF), as desired.

Further, let X = \overline{BI} \cap \overline{DF}. Then, \angle IXF = \angle IBD + \angle BDE = 180 + \angle IBC + \angle FDC = \angle CAI = \angle FEI, which implies that X lies on (AEF). A similar proof shows that \overline{CI} \cap \overline{DF} also lies on the (AEF).

Example 4 (Serbia TST 2021/2)

Let D be an arbitrary point on the side BC of triangle ABC. Points E and F are on CA and BA are such that CD=CE and BD=BF. Lines BE and CF intersect at point P. Prove that when point D varies along the line BC, PD passes through a fixed point.

We present a solution involving some trigonometry, but the bulk of the solution, is what we observed before.

Let T_A,T_B and T_C denote the intouch points of \triangle ABC. As before, we know even before we start that I is the center of (DT_AET_BT_CF). Further note that since T_CT_A \parallel DF, DT_AT_CF is a cyclic trapezoid and hence, DT_A = FT_C. Similarly, DT_A = ET_B.

We claim that the desired fixed point is the reflection of A across T_A. Now, it suffices to prove the following claim, which we do via some calculation.

Claim : Lines \overline{DP} and \overline{AT_A} are parallel.

Since we have \overline{DF} \parallel \overline{T_AF} and \overline{DE} \parallel \overline{T_AE}, we need to check that

\begin{aligned} \frac{\sin \angle T_CT_AA}{\sin \angle T_BT_AA} &= \frac{\sin \angle T_AT_CA \cdot AT_C}{\sin \angle T_AT_BA \cdot AT_B} \\ &= \frac{T_AF \cdot ET_C}{T_AE \cdot FT_B}\end{aligned}

and

\begin{aligned} \frac{\sin \angle FDP}{\sin \angle EDP} &= \frac{\sin \angle DFP \cdot FP}{\sin \angle DEP \cdot EP}\\& = \frac{\sin \angle DFC}{\sin \angle DEB} \cdot \frac{\sin \angle FEB}{\sin \angle EFC} \\ &= \frac{\sin \angle DFC}{\sin \angle EFC} \cdot \frac{\sin \angle FEB}{\sin \angle DEB} \\ &= \frac{\sin \angle FDT_A}{\sin \angle FET_B} \cdot \frac{\sin \angle EFT_C}{\sin \angle EDT_A} \\ &= \frac{T_AF \cdot ET_C}{T_AE \cdot FT_B}\end{aligned}

These are equal, so we have \overline{DP} \parallel \overline{AT_A}.

3. Swapped Equal segments to base dissection – BE=CD and CF=BD.

This condition is a bit more convoluted than the previous ones and as a result, the interpretation is also a bit messy and involved.

Theorem 5

Let ABC be a triangle, with D a point on side BC. Let E and F be points on sides AC and AF respectively such that BD=CE and CD=BF. Let \ell denote the perpendicular bisector of side BC and I the incenter. Let O_B = \overline{BI} \cap \ell and O_C = \overline{CI} \cap \ell. Then, points O_B and O_C lie on circles (BDF) and (CDE) respectively.

The strategy in approaching this condition, is to reflect D across the midpoint of side BC, to point D'. This then converts the problem to the previous configuration, with D' instead of D.

Proof : Once again, this is simply a pair of congruent triangles. Note that in \triangle O_BFB and \triangle O_CD'B we have $BD’ = CD = FB$ and \measuredangle FBO_B = \measuredangle O_BBD'. Thus, O_BF=O_BD' = O_BD, which implies that O_B is in fact the center of $(DD’F)$. Then, \measuredangle DO_BF = 2\measuredangle BD'F = \measuredangle D'BF = \measuredangle DBF
which implies that O_B indeed lies on (BDF) as desired. Similarly, it follows that O_C lies on (CDE), which finishes the proof of the claim.

Example 6 (China TST 2018/2/1)

Let ABC be a given triangle. The variable points D, E , F, respectively on sides BC, AC, AB, satisfy CD=BF and BD=CE. Point P is defined as the second intersection of (BFD) and (CED). Prove that point P lies on a fixed circle.

We start off with defining points O_B an O_C as before, and the above claim applies. Further, applying the previous section’s results to the point D' yields that the incenter I must lie on (AEF) and that \overline{BI} \cap \overline{D'E} and \overline{CI}\cap \overline{D'F} lie on (AEF).

Now, to finish off the above problem, we claim that P lies on the circle (IO_BO_C). This can be seen via a simple angle chase. First note that, \measuredangle D'FB = \measuredangle O_CO_BI. Now,

\measuredangle XIP = \measuredangle XFP = \measuredangle D'FP =

\measuredangle BFP + \measuredangle D'FB = \measuredangle IO_BP + \measuredangle O_CO_BI = \measuredangle O_CO_BP

which implies the desired result.

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Published by Lasitha Jayasinghe

4 thoughts on “Some Length Conditions on a Triangle”

  1. Good job man! These length conditions are pretty weird sometimes and they come up time to time; good thing you came up with a formal document looking at these tedious but somewhat regular configurations!

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