In a triangle, if the square of one side is equal to sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.
Given: In triangle ABC, AC2 = AB2 + BC2 To Prove: ∠B = 90°
Construction: Construct a right triangle PQR, right angled at Q,
such that PQ = AB and QR = BC
Proof: In ΔPQR, ∠Q = 90° [By construction]
D B
A C
A
B C
P
Q R
76 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume PQ2 + QR2 + = PR2 [Pythagoras theorem]
∴ AB2 + BC2 = PR2 –– (1) [By construction PQ = AB, QR = BC]
But, AB2 + BC2 = AC2 –– (2) [Given]
∴ PR2 = AC2 [from (1) and (2)]
i.e. PR = AC
In ΔABC and ΔPQR, AB = PQ [construction]
BC = QR [construction]
AC = PR [Proved above]
ΔABC ≅ ΔPQR [SSS test] ∴ ∠B = ∠Q [CPCT]
∠Q = 90° [construction] ∴ ∠B = 90°
SOLVED EXAMPLE 6.5:
1) ABC is a triangle in which AB = AC & D is any point in BC. Prove that AB2 – AD2 = BD.CD.
Sol: Construction: Draw AE ⊥ BC
Proof: AB = AC [Given]
AB2 = AE2 + BE2 [Pythagoras theorem]
AD2 = AE2 + DE2 [Pythagoras theorem]
AB2 – AD2 = AE2 + BE2 – AE2 – DE2 = BE2 – DE2
= (BE + DE) (BE – DE) = (BE + DE) BD
= (CE + DE)BD [BE = CE]
AB2 – AD2 = BD × CD
2) The perpendicular AD on the base BC of a ΔABC intersects BC at D so that DB = 3CD.
Prove that 2AB2 = 2AC2 + BC2 [CBSE 2003]
Sol: BD = 3CD
AB2 = AD2 + BD2 [Pythagoras Theorem] – I AC2 = AD2 + CD2 [Pythagoras Theorem] – II From statement (1) and (2)
AB2 = AC2 – CD2 + BD2 = AC2 + (BD2 – CD2)
= AC2 + (BD + CD) (BD – DC) = AC2 + BC [3CD – CD]
= AC2 + BC(2CD) = AC2 + BC × [4 ] 2
1BC CD=BC = AC2 + 2 BC2
∴ 2AB2 = 2AC2 + BC2
3) ABC is a right triangle, right angled at C. If p is the length of the perpendicular from C to AB and AB = c, BC = a and CA = b, then prove that, (i) pc = ab (ii) 12 12 12
b a p = + Sol: Data: ΔABC is a right angled triangle right angled at C.
CP ⊥ AB, CP = p, BC = a, AB = c, AC = b TPT: 1) pc = ab 2) 12 12 12
b a p = + Proof: A(ΔABC) = ½bh = ½pc
A(ΔABC) = ½ba [∠C = 90°]
½ab = ½pc ∴ ab = pc
According to Pythagoras theorem, C2 = a2 + b2 pc = ab ⇒ p =
ab C p c ab 1=
, 2 2 2 2
2 2 2 2
2 2
1 1 1
a b b a
b a b a
C
p = = + = + i.e. 12 12 12
a b p = + A
D C
B E
A
D C B
A
C
P B
Chapter 06: Triangles 77
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 77
UNSOLVED EXERCISE 6.5:
CW Exercise:
1) PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM. MR
2) The sides of certain triangles are given below. Determine which of them are right triangles.
i) 6 cm, 8 cm, 10 cm ii) 5 cm, 8 cm, 11 cm
3) A man goes 150 m due east and then 200 m due north. How far is he from the starting point?
4) A ladder 25 m long reaches a window of a building 20 m above the ground. Determine the distance of the foot of the ladder from the building.
5) A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its tip reaches a window 12 m above the ground. Determine the length of the ladder.
6) From a point O in the interior of a ΔABC, perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that
i) AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2 ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
7) P and Q are the mid points of the sides CA and CB respectively of a ΔABC right angled at C.
Prove that
i) 4AQ2 = 4AC2 + BC2 ii) 4BP2 = 4BC2 + AC2 iii) 4(AQ2 + BP2) = 5 AB2
8) In an isosceles triangle ABC, with AB = AC, BD is perpendicular from B to the side AC. Prove that BD2 – CD2 = 2CD AD
9) ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2 10) If ABC is an equilateral triangle of side 2a, prove that altitude AD = a 3. 11) In figure, ∠ACB = 90° and CD ⊥ AB.
Prove that, AD BD CACB =22
12) ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
13) An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1
2
1hours?
14) The perpendicular AD on the base BC of a ΔABC intersects BC at D so that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2.
15) A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D.
Prove that OB2 + OD2 = OC2 + OA2. [Hint: Through O draw a line parallel to BC]
16) Two poles of heights 6m and 11m stand vertically on a plane ground. If the distance between their feet is 12 m, determine the distance between their tops.
17) A ladder reaches a window, which is 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9m high. Find the width of the street if the length of the ladder is 15m.
18) In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9AD2 = 7AB2
19) P and Q are points on the sides CA and CB respectively of a ΔABC right angled at C. Prove that AQ2 + BP2 = AB2 + PQ2.
20) In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
HW Exercise:
1) ABC is an isosceles triangle right angled at C Prove that AB2 = 2 AC2
C
A D B
78 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume 2) A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to
the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
3) In a triangle ABC, AD is drawn perpendicular to BC. Prove that AB2 – BD2 = AC2 – CD2 5) In a triangle ABC, ∠B > ∠C, D is the mid point of BC and AE ⊥ BC. Prove that
i) AC2 = AD2 + BC.DE + 2 4
1BC (ii) AB2 = AD2 – BC.DE + 2 4
1BC (iii) AB2 + AC2 = 2AD2 + 2 2 1BC
6) ABD is a triangle in which ∠DAB = 90° and AC ⊥ BD. Prove that i) AB2 = BC × BD
ii) AC2 = BC × DC
iii) AD2 = BD × CD [CBSE–09]
7) In ΔABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5 cm and AD =
2 5
3 cm, find the length of CE.
8) BL and CM are medians of a ΔABC right angled at A. Prove that, 4(BL2 + CM2) = 5 BC2
Proof of Theorems:
Theorem 6.2: (Converse of Thales Theorem)
z If a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side.
Given:
z In ΔABC, line l intersects AB in D and AC in E, such that EC
AE DBAD =
To Prove:
z Line l || BC
Construction:
z Let line l not parallel to BC. Through D draw DF || BC
Proof:
z DF ||BC [BPT]
z FC
AF DBAD =
z But
EC AE
DBAD = [Given]
z EC
AE FCAF =
z EC
EC AE FC
FC
AF+ = + [Componendo]
A B
C D
A
F D
C B
l
E
Chapter 06: Triangles 79
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 79
z EC
AC
FCAC = ⇒ FC = EC
z But this is impossible unless F and E coincide i.e. DE is l itself
z Hence l || BC.
Theorem 6.3:
If in two triangle, corresponding angles are equal, i.e., the two triangles are equiangular, then the triangles are similar.
Given:
z Two ΔABC and ΔDEF such that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
To prove:
z ΔABC ~ ΔDEF
Construction:
z We mark point P on the line DE and Q on the line DF such that AB = DP and AC = DQ.
z We join PQ.
Proof:
z AB < DE. Thus, P lies in DE.
z In ΔABC and ΔDPQ,
z AB = DP, AC = DQ and ∠A = ∠D
z ∴ ΔABC ≅ ΔDPQ [SAS criterion of congruence]
z So, ∠B = ∠DPQ
z But, ∠B = ∠E [Given]
z ∴ ∠E = ∠DPQ
z Consequently, PQ || EF
z ∴ DE DP =
DF
DQ [Corollary to Basic Proportionally Theorem]
z i.e., DE AB =
DF
AC –– (1) [Construction]
z Similarly, DE
AB = EF
BC –– (2)
z From (1) and (2) we get, DE AB =
EF BC =
DF AC
z Since, corresponding angles are given equal, we conclude that, ΔABC ~ ΔDEF.
Theorem 6.4:
If the corresponding sides of two triangles are proportional, then they are similar.
Given:
z Two ΔABC and ΔDEF such that DE AB =
EF BC =
DF AC A
B C E F
P Q
D
80 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume
To prove:
z ΔABC ~ ΔDEF
Construction:
z We mark point P on DE and Q on DF such that AB = DP and AC = DQ. We join PQ.
Proof:
z Since DE AB =
DF
AC ,we get
z DE DP =
DF DQ
z ∴ PQ || EF [Converse of Basic Proportionally Theorem]
z So, ∠DPQ = ∠E and ∠DQP = ∠F [Corresponding angles]
z By AAA similarity, ΔDPQ ~ ΔDEF
z This gives, DE DP =
EF PQ or
DE AB =
EF
PQ –– (1)
z But, DE AB =
EF
BC –– (2) [Given]
z From (1) and (2) we get, EF PQ =
EF
BC or PQ = BC
z So, by SSS congruence criterion, ΔABC ≅ ΔDPQ
z Since, ΔDPQ ~ ΔDEF We get, ΔABC ~ ΔDEF
Theorem 6.5:
If in two triangles, one pair of corresponding sides is proportional and the included angles are equal, then the two triangles are similar.
Given:
z Two ΔABC and ΔDEF such that DE AB =
DF
AC and ∠A = ∠D
To prove:
z ΔABC ~ ΔDEF
Construction:
z We mark point P in DE and Q in DF such that AB = DP and AC = DQ. We join PQ.
Proof:
z In ΔABC and ΔDPQ, we get
z AB = DP; AC = DQ; ∠A = ∠D
z ∴ ΔABC ≅ ΔDPQ [SAS criterion of congruence] –– (1)
z Now, DE AB =
DF
AC [Given]
z DE DP =
DF
DQ [By construction]
A
B C E F
P Q
D
A
B C E F
P Q
D
Chapter 06: Triangles 81
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 81 z So, PQ || EF [Converse of Basic Proportionally Theorem]
z ∴ ∠DPQ = ∠E and ∠DQP = ∠F
z Consequently, by AA similarity, ΔDPQ ~ ΔDEF
z Hence, ΔABC ~ ΔDEF [By (1)]