1) Exercise 1
In the diagram, ∠ABC = ∠AED, AD = 3, DB = 2 and AE = 2. Determine the length
of EC.
Solution:
First, we show that ∆AED and ∆ABC are similar. Since ∠DAE = ∠BAC and ∠ABC =
CA
= AB
. By cross multiplying,
∠AED, we have that ∆AED is similar to ∆ABC. Thus, DA
AE
CA =
Hence, EC = CA − AE =
15
2
DA · AB
3
15
= · (3 + 2) = .
AE
2
2
−2=
11
.
2
1
2) In triangle ABC, D is a point on BC. Further, AB = 35, BD = 11 and AD = AC = 31.
Determine the length of DC.
Solution: Construct the Perpendicular as shown
Now, by the Pythagorean Theorem on ∆ABE, we see that
(AB)2 = (BE)2 + (AE)2
352 = (11 + DE)2 + (AE)2
352 = 112 + 22 · DE + (DE)2 + (AE)2
Now, by the Pythagorean Theorem on ∆ADE, we see that
(AD)2 = (DE)2 + (AE)2
312 = (DE)2 + (AE)2
Subtracting these two equations gives
352 − 312 = 112 + 22 · DE
Either using a calculator of factoring as follows, we see that
352 − 312 = 112 + 22 · DE
(35 − 31)(35 + 31) = 11(11 + 2 · DE)
(4)(66) = 11(11 + 2 · DE)
(4)(6) = 11 + 2 · DE
24 − 11 = 2 · DE
13 = DC
2
Since 2DE = DC
3) In triangle ABC, AC = AB = 25 and BC = 40. Point D is a point chosen on BC. From
D, perpendiculars are drawn to meet AC at E and AB at F . Determine the value of
DE + DF .
Solution: As shown in the diagram, draw the perpendicular bisector of CB which intersects A and M, the midpoint of BC. Also connect A and D
Thus, M C = 20 and hence by the Pythagorean Theorem,
(AC)2 = (CM )2 + (AM )2
252 − 202 = (AM )2
(25 − 20)(25 + 20) = (AM )2
5 · 45 = (AM )2
52 · 92 = (AM )2
15 = AM
Hence, the area of ∆ABC is
see that
AM ·BC
2
= 300. Now, denoting area by absolute values, we
300 = |∆ABC| = |∆ADC| + |∆ADB|
AC · ED AB · F D
300 =
+
2
2
25 · ED 25 · F D
300 =
+
2
2
600 = 25(ED + F D)
24 = ED + F D
3
4) Triangle ABC is an isosceles triangle in which AB = AC = 10 and BC = 12. The points
S and R are on BC such that BS : SR : RB = 1 : 2 : 1. The midpoints of AB and AC
are P and Q respectively. Perpendiculars are drawn from P and R to meet SQ at M and
N respectively. What is the length of M N ?
Solution: From the problem statement, P and Q are midpoints hence AP = P B =
AQ = AC = 5. Let X be the midpoint of BC. Since the triangle ABC is isosceles, we see
that AM is the perpendicular bisector of BC. Similarly, AY is the perpendicular bisector
for ∆AP Q. Since the sides are in a 1 : 2 : 1 ratio, we see that BS = SX = XR = RC = 3.
First, we claim that QR ⊥ BC. Draw the triangle below as follows.
Then ∆AQY is similar to ∆ACM (They share ∠CAX = ∠QAY and both have a right
angle). Hence
AQ
AC
5
10
=
implying
=
AY
XC
AY
8
Hence AY = 4. Similarly, AQ = 3. By the Pythagorean Theorem in ∆AM C, we see that
(AC)2 = (AX)2 + (M C)2
(10)2 = (AX)2 + 62
100 − 36 = (AX)2
8 = AX
4
Hence, XY = 4 and by the Pythagorean Theorem again, we see that QX = 5. Hence,
∆QM C is isosceles and thus, QR must be perpendicular since it is the bisector of CX.
Now, we can argue as above to show that P S is perpendicular to BC. By the Pythagorean
Theorem on ∆QRC, we see that
(QC)2 = (QR)2 + (RC)2
(5)2 = (QR)2 + 32
25 − 9 = (QR)2
4 = QR
and similarly, P S = 4. Now, as P S and QR are parallel, we see that ∠P SM = ∠SQR.
Hence ∆P SM is congruent to ∆RQN (angles are equal and they share a side length size).
Thus SM = N Q. Since ∠QRS is a right angle, we see that ∠SRN = ∠N QR. Further,
∠RSN = ∠QRN . Hence ∆RN Q which is similar to ∆SRQ. This gives
QR
QN
=
QR
QS
Thus, QN · QS = 16. Now, by Pythagorean Theorem again on ∆QRS, we see that
(QS)2 = (QR)2 + (SR)2
(QS)2 = 42 + 62
(QS)2 = 16 + 36
√
QS = 52
√
QS = 2 13
Thus, QN =
16
√
2 13
=
√8 .
13
Now, QS = SM + M N + N Q = 2QN + M N and so
√
8
26 − 16
10
M N = QS − 2QN = 2 13 − 2 · √ = √
=√
13
13
13
completing the problem.
5
5) The lengths of the diagonals AD and BC in rhombus ABCD are 6 and 8 respectively.
Triangle AXY is equilateral and line XY is parallel to diagonal BC. Determine the
length of the altitude of triangle AXY .
Solution: Construct the picture as shown:
Our goal is to find the length of AF . Lines AC and BD are the diagonals. Since the
diagonals of a rhombus bisect each other, we see that EC = AE = 6/2 = 3 and DE =
BE = 8/2 = 4. Note also that diagonals of a rhombus meet at right angles. Now, since
DB and XY are parallel, ∠AF Y = ∠AEB = 90◦ . Thus, since ∠F Y A = 60◦ , we see that
∠Y AF = 30◦ . Thus, ∆Y AF is a magic triangle and so
√
3
AF
=
FY
1
√
√
Giving AF = 3 · F Y and so AF = 3 + EF = 3 · F Y . Next, we note that ∆CEB and
∆CF Y are similar since they share ∠Y CF = ∠BCE and ∠CEB = ∠CF Y . Thus, by
similar triangles, we see that
FC
CE
3
=
= .
FY
BE
4
Hence 4F C = 3F Y . Now,
√ F C = 3 − EF and thus, 12 − 4EF = 3F Y . Solving for F Y
by substituting EF = 3 · F Y − 3 gives
√
12 − 4( 3 · F Y − 3) = 3F Y
√
12 − 4 3 · F Y + 12 = 3F Y
√
24 = 3F Y + 4 3F Y
√
24 = (3 + 4 3)F Y
24
√ = FY
3+4 3
√
√
24 √3
Thus, AF = 3 · F Y = 3+4
completing the question.
3
6
6) In the diagram, ABCD is a rhombus with K the midpoint of DC and L the midpoint
of BC. Segments DL and BK intersect at M . Determine the fraction of the area of
quadrilateral KM LC is of the area of the rhombus ABCD.
Solution: Denote areas by absolute values. Join the diagonal DB and points M C as is
done in the following diagram
Now, Since K is the midpoint of DC, we see that |∆M DK| = |∆M KC|. Similarly,
|∆M LC| = |∆M LB|. Looking at ∆DBC, we see that ∆BDK| = |∆BKC|. Hence, we
have that
|∆BDM | + |∆M DK| = |KM LC| + |∆BM L|
Similarly with ∆DBL and ∆DLC, we see that
|∆BDM | + |∆BM L| = |KM LC| + |∆M DK|
Subtracting these two gives
|∆M DK| − |∆BM L| = |∆BM L| − |∆M DK|
Which gives that 2|∆BM L| = 2|∆M DK| and hence |∆BM L| = |∆M DK|. This gives
us that
|KM LC| = |∆M KC| + |∆M LC| = |∆M DK| + |∆BM L| = 2|∆M DK|
Thus, in the above, we see that ∆BDM | = |KM LC| = 2|∆M DK|. Since ∆DBC is half
the rhombus, we see that
|ABCD|/2 = |∆DBC| = |∆BDM | + |KM LC| + |∆M DK| + |∆BLM |
= 6|∆M DK|
= 3|KM LC|
and hence KM LC is one sixth of the area of the rhombus.
7
7) Exercise 3:
In triangle ABC, point D is on AB such that AD is twice as long as DB and E is a point
on BC such that BE is twice as long as BC. If the area of triangle ABC is 90 units
squared, what is the area of triangle ADE in units squared?
Solution: From the problem statement, we see that BE = 2BC. Note that ∆ABE and
∆AEC have the same heights and so their areas are in proportion with their bases. Thus,
denoting area by absolute values, |∆ABE| = 2|∆AEC|. The problem also tells us that
(suppressing units throughout)
90 = |∆ABC| = |∆ABE| + |∆AEC| = 3|∆AEC|
and hence |∆AEC| = 30. Also from the problem statement, we see that AD = 2DB.
Note that ∆AED and ∆DEB have the same heights and so their areas are in proportion
with their bases. Thus, denoting area by absolute values, |∆AED| = 2|∆DEB|. Since
60 = 2|∆AEC| = |∆ABE| = |∆AED| + |∆DEB| = 3|∆DEB|
we see that |∆DEB| = 20 and hence |∆AED| = 2|∆DEB| = 40.
8
8) Exercise 2
Square ABCD has an area of 4. The point E is the midpoint of AB. Similarly, F, G, H
and I are the midpoints of DE, CF , DG and CH respectively. What is the area of
triangle IDC?
Solution: As usual, denote area by absolute values. Connect EC to form an isosceles
triangle DEC. Now, since F is the midpoint of DE, we see that |∆DF C| = 0.5|∆DEC|
since the median cuts the area in half. Similarly, |∆DGC| = 0.5|∆DF C|, . |∆DHC| =
0.5|∆DGC|, |∆DIC| = 0.5|∆DHC|. Combining this gives |∆DIC| = (0.5)4 |∆DEC| =
1
· |∆DEC|. Since the height of ∆DEC is 2 and the base is 2, it’s area is 2 and hence
16
2
= 81 .
∆DIC| = 16
9
9) In the diagram, AB = AC = 12cm and AE = AD = 8cm. The area of quadrilateral
AEF D is 8cm2 . What is the area of triangle ABC in square centimetres?
Solution: We suppress units until the end of this argument. Construct segment AF as
shown.
Now, ∆AEF and ∆EF B have the same height and their bases are in a 2 : 1 proportion
with each other. Further, AF cuts the area of quadrilateral AEF D in half by symmetry
so ∆AEF has area 4. Hence, the area of ∆EF B is half the area of ∆AEF which gives
the area of ∆EF B to be 2. By symmetry, ∆DF C also has area 2. Lastly, using ∆AEC
which has area 8 + 2 = 10 (adding the area of quadrilateral to the triangle DF C) and
using ∆ECB which has area 2 + |∆F CB| (here the absolute value signs denote area),
we see that since the heights of these triangles are the same and the side lengths are in
and thus |∆F CB| = 3.
a 2 : 1 ration with each other, we have that 2 + |∆F CB| = 10
2
Adding the three triangle areas and the quadrilateral area gives 8 + 2 + 2 + 3 = 15 square
centimetres as the area.
10
10) The diagram shows a square ABCD with unit length. Triangle ADE is equilateral. The
diagonal AC of square ABCD intersects line segment DE at the point F . What is the
area of triangle AF D?
Solution: Begin by constructing the following line segments on the diagram
Our goal is to find |∆F DC|, the area of ∆F DC. This is given by |∆F DC| = 21 F H ·DC =
FH
. Now, F GDH is a rectangle and hence GD = F H. Since AC is a diagonal of a square,
2
∠HCF = 45◦ . Since ∆F HC is a magic triangle, we see that GD = F H = HC. Thus,
DH = DC − HC = 1 − GD. Since F GDH is a rectangle, once again we have that
F G = DH = 1 − GD. Since ∆F GD is a 30◦ : 60◦ : 90◦ triangle, we see that
GD
1
=√
FG
3
GD
1
=√
1 − GD
3
√
3 · GD = 1 − GD
√
( 3 + 1)GD = 1
Solving gives GD =
√
3−1
4
√1
3+1
√
or GD =
3−1
.
2
Since GD = F H, we see that |∆F DC| =
completing the question.
11
FH
2
=
11) Below, F is the midpoint of AE, AE = 21 ED, AB = 9 and BC = 3. If the area of
quadrilateral BEDC is 72, then what is the area of triangle BF E?
Solution: Join BD. Denoting area by absolute values, we see that |∆BF E| = |∆ABF |
since the triangles have the same base and height. Since DE = 2AE, we see that
|∆BED| = 2|∆BAE| = 2(|∆BF E| + |∆ABF |) = 2(|∆BF E| + |∆BF E|) = 4|∆BF E|
Now, Since AB : BC is 3 : 1, we see that |∆ABD| = 3|∆BCD|. Since
|∆ABD| = |∆BF E|+|∆ABF |+|∆BED| = |∆BF E|+|∆BF E|+4|∆BF E| = 6|∆BF E|
Combining gives |∆BCD| = 2|∆BF E|. Thus,
72 = |BEDC| = |∆BED| + |∆BCD| = 4|∆BF E| + 2|∆BF E| = 6|∆BF E|
Hence, |∆BF E| = 12.
12
12) In the diagram, triangle ABC is equilateral with sides of length 2. Line segments CD
and EB are medians and F GHI is a square. Determine the ratio of the area of square
F GHI to triangle ABC.
Solution: Draw the altitude AM which is also the perpendicular bisector of BC. Let
M G = F G = n so that the square has side length 2n.
From the problem statement, we know that M C = BC/2 = 1. Hence GC = 1 − n. Now,
triangle HGC is a magic triangle since the angles are 30◦ : 60◦ : 90◦ . Hence, we see that
2n
HG
1
=
=√
1−n
GC
3
√
This gives 2 3n = 1 − n and so n = 2√13+1 . Hence, the area of the square F GHI is
√
√
4√
1
1
4n2 = 13+4
.
The
area
of
the
triangle
ABC
is
·
AM
·
BC
=
3
·
2
=
3 (note that
2
2
3
∆ABM is another magic triangle) and thus, the ratio of the area of this square to triangle
ABC is
4√
4
13+4 3
√
√
=
3
12 + 13 3
13