50 AMC LECTURES Lecture 19 Eight More Methods To Draw

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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
BASIC KNOWLEDGE

In this lecture, we introduce eight more commonly used methods to draw auxiliary lines

1. Double the length of the median of a triangle.

In triangle ABC, AD is the median on side BC. If we extend AD to E
such that DE = AD and connect CE, we get two congruent triangles
CDE and BDA.

Example 1: In ABC, AD is the median, BE and AC meet at E. BE and AD meet at F. If
AE = EF, show AC = BF.

Proof:

Extending AD to H such that DH = AD.

Since BD = CD, BDH =ADC, then ACD  HBD, AC =
BH, DAC =H.

We are given that AE = EF, so AFE = EAF = BFH.
Therefore in BFH, BFH =H, BF = BH = AC .

2. Draw the height of the figure (especially when area calculation is involved).

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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
Example 2: (1972 AMC) A triangle has angles of 30 and 45. If the side opposite the
45 angle has length 8, then the side opposite the 30 angle has length
(A) 4 (B) 4 2 (C) 4 3 (D) 4 6 (E) 6

Solution: (B).
Let s denote the length of the required side (see figure).
Then the altitude to the longest side, opposite the 30 angle,
8
has length 4 and is one leg of an isosceles right triangle
2
with hypotenuse s, which therefore has length 4 2.

Example 3: (1973 AMC) Two congruent 30 – 60 – 90 triangles
are placed so that they overlap partly and their hypotenuses
coincide. If the hypotenuse of each triangle is 12, the area common
to both triangles is
(A) 6 3 (B) 8 3 (C)9 3 ( D) 12 3 (E) 24

Solution: (D).
In the adjoining figure MV is an altitude of ∆AMV (a 30 – 60 – 90 triangle), and MV
1
has length2 3. The required area is, therefore, area ABV  (AB)(MV)
2
1
 122 3 12 3.
2

3. Draw the diagonals of a parallelogram.

ABCD is a parallelogram. AC and BD are the diagonals.
AC2 BD2  AB2 BC2 CD2 DA2

Proof:
Draw DE  AB, CF  AB.
By the Pythagorean Theorem,
AC2  AF2 CF2 (ABBF)2 (BC2 BF2) AB2 BC2 2ABBF (1)
BD2 BE2 DE2 (AB AE)2 (AD2  AE2) AB2  AD2 2ABAE (2)
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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
Since ∆ADE  ∆BCF (AD = BC, DE = CF, AED = BFC = 90), AE = BF.
(1) + (2): AC2 BD2  AB2 BC2 CD2 DA2.

Theorem: A diagonal of a parallelogram divides the parallelogram into two congruent
triangles.

AED  CEB and AEB  CED.

Theorem: The diagonals of a parallelogram bisect each other. Also
converse.

AE = EC and DE = EB.

Example 4: In triangle ABC, if AB = c, AC = b, BC = a and O is the
midpoint of AC. Find m , the length of the median BO.
b

Solution:

Extending BO to D such that BO = OD. Connect AD and CD. Since AC
and BD bisect each other, they are two diagonals of a parallelogram, that is, ABCD is a
parallelogram.
AC BD AC
Therefore 2(AB2 BC2) AC2 BD2 4( )2 4( )2 4( )2 4BO2
2 2 2
AC b
 AB2 BC2 2[BO2 ( )2]  c2 a2 2[m2 ( )2]
2 b 2
1
 m  2a2 2c2 b2 .
b 2
This is the formula to calculate the median of a triangle if three sides are known.
1 1
Similarly, we can have: m  2b2 2c2 a2 and m  2a2 2b2 c2 .
a 2 c 2
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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
Example 5: DB is the diagonal of parallelogram ABCD. EF//DB and
meets BC at E and DC at F, respectively, as shown in the figure.
Show that triangle ABE and triangle ADF have the same areas.

Solution:
Draw AC, the second diagonal of ABCD.
S BE S DF
ABE  ; ADF  .
S BC S DC
ABC ADC
BE DF S S
Since EF ⁄⁄DB,   ABE  ADF
BC DC S S
ABC ADC
We also know that S S (∆ABC  ∆CDA).
ABC ADC
ThereforeS S .
ABE ADF

4. Translating a diagonal or a leg of trapezoid to form a parallelogram.

(1). In trapezoid ABCD, AB//DC. Draw BE//AC and
meet the extension of DC at E.

We get:
BE = AC, AB = CE, DE = DC + CE = DC + AB

When AD = BC, we get BD = AC = BE.

(2). In trapezoid ABCD, AB//DC. Draw CF//AD to meet AB at
F.
We get: AD = FC and AF = DC.

Example 6: (1970 AMC) In the accompanying figure, segments AB and CD are
parallel, the measure of angle D is twice that of angle B, and the
measures of segments AD and CD are a and b respectively. Then
the measure of AB is equal to
1 3 3 1
(A) a2b (B) b a (C) 2a – b (D) 4b a (E) a + b.
2 2 4 2

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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
Solution: (E).
Let the bisector of D intersect AB at P (see figure). Then the alternate interior angles
APD and PDC as well as ADP are equal to angle B, so that ∆APD
is isosceles with equal angles at P and D. This makes AP = AD = a.
Since PBCD is a parallelogram, we have PB = DC = b; so AB = AP
+ PB = a + b.

Example 7: In a convex quadrilateral ABCD, AD  BC. Show that
ACBD if AC2 + BD2 = (AD + BC)2.

Solution: Draw DE//AC and meets the extension of BC at E. Then
∆CED  ∆DAC. DE = AC, CE = AD.
In ∆BDE, BE = BC + CE = BC + AD. DE = AC.
Since AC2 + BD2 = (AD + BC)2, or DE2 + BD2 = BE2, by the
converse of Pythagorean theorem, BDE = 90.
Therefore BDDE. We also know that AC ⁄⁄ DE. So ACBD.

5. Draw the perpendicular to chord through the center of a circle

O is the center of the circle. Draw OC  AB.
We have AC = CB and AD = DB

Theorem: A line perpendicular to a chord of a circle and containing the center of the
circle, bisects the chord and its major and minor arcs.
Theorem: The perpendicular bisector of a chord of a circle contains the center of the
circle.

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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
Example 8: (1973 AMC) A chord which is the perpendicular bisector of radius of length
12 in a circle, has length
(A) 3 3 (B) 27 (C) 6 3 (D) 12 3 (E) none of these

Solution: (D).
Let O denote the center of the circle, and let OR and AB be the radius and the
chord which are perpendicular bisectors of each other at M. Applying the
Pythagorean theorem to right triangle OMA yields (AM)2 = (OA)2 – (OM)2 =
1221 – 62 = 108, AM = 6 3.
Thus the required chord has length 12 3.

6. Draw the inscribed angle of the diameter

Theorem: An angle inscribed in a semicircle is a right angle.

Theorem: The measure of an inscribed angle equals one-half the measure
180
of its intercepted arc. C  90
2

Example 9: (1995 AMC) In the figure, AB and CD are diameters of the
circle with center O, AB  CD, and chord DF intersects AB at E. If DE =
6 and EF = 2, then the area of the circle is
47 49
(A) 23 (B)  (C) 24 (D)  (E) 25
2 2

Solution: (C).

Draw segment FC. Angle CFD is a right angle since arc CFD is a
semicircle. Then right triangles DOE and DFC are similar, so
DO DE
 .
DF DC
Let DO = r and DC = 2r. Substituting, we have
r 6
  2r2 48  r2 24.
8 2r
Then the area of the circle is r2 = 24.
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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
Example 10: AB is the diameter of circle O. C is a point on the circumference. P is a
point on the circumference PF is perpendicular to AB. PF meets AC at E,
AB at D and the extension of BC at F. Show thatDP2 DEDF.

Solution:

Connect PA, and PB. APB = 90. ACB = 90.
We have: DP2  ADDB. Now we only need to prove
ADDBDEDF.
We also see that .∆ADE~∆FDB.
We know that F = EAD, and ADE =FDB = 90.
AD DF
Therefore ∆ADE~∆FDB    ADDBDEDF.
DE DB

7. When two circles are intersecting, draw the common chords or connect the centers.

O O is the perpendicular bisector of EF. O C  DC.
1 2 2

Theorem: Any point on the perpendicular bisector of a line
segment is equidistant from the endpoints of the line
segment. Two points equidistant from the endpoints of a
line segment, determine the perpendicular bisector of the
line segment.

Example 11: (1966 AMC) The length of the common chord of two intersecting circles is
16 feet. If the radii are 10 feet and 17 feet, a possible value for the distance between the
centers of the circles, expressed in feet, is:
(A) 27 (B) 21 (C) 389 (D) 15 (E) undetermined
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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
Solution: (B).
Denote the common chore by AB, its midpoint by P, and the
centers of the smaller and larger circles by O and O; OO is
perpendicular to AB and passes through P.
The Pythagorean Theorem applied to right triangles OPA and
OPA now yields
OP2 OA2 AP2 102 82 36, OP6,
And OP2 OA2 AP2 172 82 225, OP15. 15 + 6 = 21.

8. When a figure looks like a part of the other figure, draw the original figure.

Example 12: (1968 AMC) Let side AD of convex quadrilateral ABCD be extended
through D, and let side BC be extended through C, to meet in point E. let S represent the
degree-sum of angles CDE and DCE, and let S represent the degree-sum of angles BAD
S
and ABC. If r = , then:
S
(A) r = 1 sometimes, r > 1 sometimes (B) r = 1 sometimes, r < 1 sometimes
(C) 0 < r < 1 (D) r > 1 (E) r = 1

Solution: (E).
Since we extended AD and BC to meet at E, Triangle ABE is the
original figure.
We know that the sum of the angles in a triangle is 180(see diagram
on right), so E + CDE +DCE = E + S = 180 in ∆EDC.
And E + BAD + ABC = E + S = 180 in ∆EAB.
S
Hence S = S = 180 – E, so that r = = 1.
S

Example 13: (1972 AMC) Let ABCD be a trapezoid with the measure of base AB twice
that of base DC, and let E be the point of intersection of the diagonals. If the measure of
diagonal AC is 11, then that of segment EC is equal to
2 3 1
(A) 3 (B) 3 (C) 4 (D) 3 (E) 3
3 4 2
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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
Solution: (A).
Extend sides AD and BC to meet at V. Triangle ABV is the
original figure.
Then AC and BD are medians from vertices A and B of ∆ABV
meeting at point E, which divides the length of each in the
ratio 2 : 1. This means that
1 11 2
EC  AC  3 .
3 3 3

Example 14: (1999 AMC #23) The equiangular convex hexagon ABCDEF has AB = 1,
BC = 4, CD = 2, and DE = 4. The area of the hexagon is
15 39 39
(A) 3 (B) 9 3 (C)16 (D) 3 (E) 3
2 4 4

Solution:
Extend FA and CB to meet at X, BC and ED to meet at Y , and DE
and AF to meet at Z. Triangle XYZ is the original figure. The interior
angles of the hexagon are 120. Thus the triangles XYZ, ABX, CDY,
and EFZ are equilateral. Since AB = 1, BX = 1. Since CD = 2, CY = 2.
Thus XY = 7 and YZ = 7. Since YD = 2 and DE = 4, EZ = 1. The area
of the hexagon can be found by subtracting the areas of the three
small triangles from the area of the large triangle:
3 3 3 3 43 3
72( )12( )22( )12( ) .
4 4 4 4 4

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50 AMC LECTURES Lecture 19 Eight More Methods To Draw Auxiliary Lines
PROBLEMS

Problem 1. (1980 AMC) Sides AB, BC, CD and DA of convex quadrilateral ABCD have
lengths 3, 4, 12, and 13, respectively; and CBA is a right angle. The area of the
quadrilateral is
(A) 32 (B) 36 (C) 39 (D) 42 (E) 48

Problem 2. (1967 AMC) In quadrilateral ABCD with diagonals AC and BD intersecting
at O, BO = 4, OD = 6, AO = 8, OC = 3, and AB = 6. The length
of AD is
(A) 9 (B) 10 (C) 6 3 (D) 8 2 (E) 166

Problem 3. In ABC, C=90. D is the middle point of BC.
DEAB at E. Prove BC2  2BEAB

Problem 4: (2003 AMC 12 B) In rectangle ABCD, AB = 5 and BC = 3. Points F and G
are on CD so that DF = 1 and GC = 2. Lines AF and BG intersect at E.
Find the area of AEB.
21 25
(A) 10 (B) (C) 12 (D) (E) 15
2 2

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