# 1) Exercise 1 In the diagram, ABC = AED, AD = 3, DB = 2 and AE = 2. Determine the length of EC. Solution:

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1 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 = AED, we have that AED is similar to ABC. Thus, DA. By cross multiplying, CA = DA AB AE Hence, EC = CA AE = = = 3 2 = CA AE AB (3 + 2) =

2 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) = (11 + DE) 2 + (AE) = DE + (DE) 2 + (AE) 2 Now, by the Pythagorean Theorem on ADE, we see that (AD) 2 = (DE) 2 + (AE) = (DE) 2 + (AE) 2 Subtracting these two equations gives = DE Either using a calculator of factoring as follows, we see that = DE (35 31)( ) = 11( DE) (4)(66) = 11( DE) (4)(6) = DE = 2 DE 13 = DC Since 2DE = DC 2

3 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, MC = 20 and hence by the Pythagorean Theorem, (AC) 2 = (CM) 2 + (AM) = (AM) 2 (25 20)( ) = (AM) = (AM) = (AM) 2 15 = AM Hence, the area of ABC is AM BC 2 = 300. Now, denoting area by absolute values, we see that 300 = ABC = ADC + ADB AC ED 300 = + AB F D ED 300 = + 25 F D = 25(ED + F D) 24 = ED + F D 3

4 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 MN? 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 AY = AC 5 implying XC AY = 10 8 Hence AY = 4. Similarly, AQ = 3. By the Pythagorean Theorem in AMC, we see that (AC) 2 = (AX) 2 + (MC) 2 (10) 2 = (AX) = (AX) 2 8 = AX 4

5 Hence, XY = 4 and by the Pythagorean Theorem again, we see that QX = 5. Hence, QMC 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) = (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 = NQ. Since QRS is a right angle, we see that SRN = NQR. Further, RSN = QRN. Hence RNQ which is similar to SRQ. This gives QN QR = QR QS Thus, QN QS = 16. Now, by Pythagorean Theorem again on QRS, we see that (QS) 2 = (QR) 2 + (SR) 2 (QS) 2 = (QS) 2 = QS = 52 QS = 2 13 Thus, QN = = Now, QS = SM + MN + NQ = 2QN + MN and so MN = QS 2QN = = = completing the problem. 5

6 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 AF F Y = 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 F C F Y = CE BE = 3 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 Thus, AF = 3 F Y = ( 3 F Y 3) = 3F Y F Y + 12 = 3F Y 1 24 = 3F Y + 4 3F Y 24 = ( )F Y = F Y completing the question. 6

7 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 KMLC is of the area of the rhombus ABCD. Solution: Denote areas by absolute values. Join the diagonal DB and points MC as is done in the following diagram Now, Since K is the midpoint of DC, we see that MDK = MKC. Similarly, MLC = MLB. Looking at DBC, we see that BDK = BKC. Hence, we have that BDM + MDK = KMLC + BML Similarly with DBL and DLC, we see that Subtracting these two gives BDM + BML = KMLC + MDK MDK BML = BML MDK Which gives that 2 BML = 2 MDK and hence BML = MDK. This gives us that KMLC = MKC + MLC = MDK + BML = 2 MDK Thus, in the above, we see that BDM = KMLC = 2 MDK. Since DBC is half the rhombus, we see that ABCD /2 = DBC = BDM + KMLC + MDK + BLM = 6 MDK = 3 KMLC and hence KMLC is one sixth of the area of the rhombus. 7

8 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

9 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 DIC = 2 =

10 9) In the diagram, AB = AC = 12cm and AE = AD = 8cm. The area of quadrilateral AEF D is 8cm 2. 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 = 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 a 2 : 1 ration with each other, we have that 2 + F CB = 10 and thus F CB = 3. 2 Adding the three triangle areas and the quadrilateral area gives = 15 square centimetres as the area. 10

11 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 = 1F H DC = 2 F H. 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 F G = 1 3 GD 1 GD = GD = 1 GD ( 3 + 1)GD = 1 Solving gives GD = or GD = 3 1. Since GD = F H, we see that F DC = F H = completing the question. 4 11

12 11) Below, F is the midpoint of AE, AE = 1 ED, AB = 9 and BC = 3. If the area of 2 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 =

13 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 MG = F G = n so that the square has side length 2n. From the problem statement, we know that MC = 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 1 n = HG GC = 1 3 This gives 2 3n = 1 n and so n = 1 2. Hence, the area of the square F GHI is 3+1 4n 2 = The area of the triangle ABC is 1 AM BC = = 3 (note that ABM is another magic triangle) and thus, the ratio of the area of this square to triangle ABC is =

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