Denise Chua (August 26, 2025)
Session 2, Unit 2: Angle Bisector Theorem
1 Main Learnings
1.1 Angle Bisector + Altitude = Congruent Triangles
If a line is an angle bisector and altitude of a triangle, then the two triangles formed
by the split of the line are congruent by ASA. Since the Angle Bisector Theorem (ABT)
mainly relies on knowing certain side lengths of triangles with angle bisectors, this
can help solve for desired lengths.
ABT: 2016 HMMT November Theme #3
The three points A, B, C form a triangle. AB = 4, BC = 5, AC = 6. Let the angle
bisector of ∠A intersect side BC at D. Let the foot of the perpendicular from B
to the angle bisector of ∠A be E. Let the line through E parallel to AC meet BC
at F . Compute DF .
We see AE is both an angle bisector and altitude of △ABG. So, ASA congruence
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BE
BF
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yields. △AEB ∼
= △AEG, and BE
BG = 2 . Note △BEF ∼ △BGC, so BG = BC = 2 helps
us solve for BF. Finally, subtract BF by BD to obtain the desired DF.
ABT: From AoPS
Given △ABC where D and E are on AC such that ∠ABD = ∠DBE = ∠EBC
and D is between A and E, if DE = 4, EC = 5,
√ and BD = 12, find BE × AD.
Express your answer in simplest radical form (a b).
ABT yields BC = 15, revealing that △CDB √
is a 3 − 4 − 5 triangle. Thus, we use
Pythagorean Theorem (PT) to obtain BE = 4 10. Secondly, we see BD is an angle
bisector and altitude, so △ADB ∼
= △EDB. Hence, AD = DE = 4, and we calculate
BE × AD.
Overall, this shows how mentioning “angle bisectors” in a problem can lead to congruent
triangles. This, combined with ABT, gives us a better understanding of the side lengths
and angles of the figure.
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Denise Chua (August 26, 2025)
Session 2, Unit 2: Angle Bisector Theorem
1.2 Coordinate Geometry Bashing
Most problems can be solved with Euclidean geometry. However, when that’s hard to
see, coordinate geometry becomes the second option. Here’s a problem that can test
one’s ability to bash with coordinates:
ABT: 1990 AIME #7
A triangle has vertices P = (−8, 5), Q = (−15, −19), and R = (1, −7). The
equation of the bisector of ∠P can be written in the form ax + 2y + c = 0. Find
a + c.
Let X be the foot of the angle bisector from P to QR. First, we use the distance formula
on all three sides. Next, use ABT to determine QX
XR . With this ratio, we apply the internal
section formula to solve for the coordinates of X. Finally, use the two-point form to solve
for the line from P to X.
Even though bashing doesn’t use much clever insight and isn’t required for most problems,
it could still be helpful in last resort situations during math contest examinations.
2 Feedback
2.1 Use of ABT is Too Obvious
Most, if not all problems, mention the word angle bisector. From here on, most
problems can be solved by just using ABT as the first step with pre-existing knowledge
on congruent triangles, similar triangles, area ratios, etc. Here’s an external problem
that disguises the use of ABT well:
Self-found: 2022 IWYMIC Problem #4
In the diagram below, let P be a point on the circumcircle of the equilateral
triangle ABC such that P B = 24cm and P C = 8cm. If AP and BC intersect at D,
then find the length, in cm, of PD.
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Denise Chua (August 26, 2025)
Session 2, Unit 2: Angle Bisector Theorem
We’re motivated to angle chase because A, B, P, C are concyclic. Note that ∠ABC =
∠ACB = 60◦ . Thus, ∠AP C = ∠AP B = 60◦ too. So, PD is an angle bisector of △BP C.
Now, we can use ABT, giving us BD = 3x and DC = x.
Because △ABC is equilateral, AB = BD + DC = 4x. Now, to solve for P D, we can try
to use the side ratios we obtained through similar triangles. Note △ABD ∼ △CP D by
AA similarity:
AB
CP
4x
8
=
=⇒
=
BD
PD
3x
PD
This solves for P D. Overall, adding more problems with angle chasing or extra steps
helps students detect the use of ABT beyond seeing the words “angle bisector.”
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