Example-1 :- If a line intersects sides AB and AC of a Δ ABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC (see figure).
Given that : DE || BC.
Prove that : AD/AB = AE/EC
Proof : We have DE || BC (Given)
so, AD/DB = AE/EC (Theorem-1)
DB/AD = EC/AE
Both sides adding by 1, we get
DB/AD + 1 = EC/AE + 1
(DB + AD)/AD = (EC + AE)/AE
AB/AD = AC/AE or
AD/AB = AE/AC
Example-2 :- ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Figure). Show that AE/ED = BF/FC.
Given that : ABCD is a trapezium with AB || DC and EF || AB. Prove that : AE/ED = BF/FC. Construction : Let us join AC to intersect EF at G.Proof : We have AB || DC and EF || AB (Given) so, EF || DC (Lines parallel to the same line are parallel to each other) Now, in Δ ADC, EG || DC (As EF || DC) So, AE/ED = AG/GC ......(i) (Theorem-1) Similarly, from Δ CAB, CG/AG = CF/BF or AG/GC = BF/FC .......(ii) Therefore, from (i) and (ii) AE/ED = BF/FC
Example-3 :- In figure, PS/SQ = PT/TR and ∠ PST = ∠ PRT. Prove that PQR is an isosceles triangle.
Given that : PS/SQ = PT/TR and ∠ PST = ∠ PRQ.
Prove that : PQR is an isosceles triangle.
Proof : We have PS/SQ = PT/TR (Given)
so, ST || QR (Theorem-2)
Therefore, ∠ PST = ∠ PQR....(i) (Corresponding angles)
Also, we have
∠ PST = ∠ PRQ ......(ii) (Given)
So, ∠ PRQ = ∠ PQR [From (i) and (ii)]
Therefore, PQ = PR (Sides opposite the equal angles)
i.e., PQR is an isosceles triangle.
Example-4 :- In Figure, if PQ || RS, prove that Δ POQ ~ Δ SOR.
Given that : PQ || RS
Prove that : Δ POQ ~ Δ SOR
Proof : We have PQ || RS (Given)
So, ∠ P = ∠ S (Alternate angles)
and ∠ Q = ∠ R
Also, ∠ POQ = ∠ SOR (Vertically opposite angles)
Therefore, Δ POQ ~ Δ SOR (AAA similarity criterion)
Example-5 :- Observe Figure and then find ∠ P.
In Δ ABC and Δ PQR,
AB/RQ = 3.8/7.6 = 1/2
BC/QP = 6/12 = 1/2
CA/PR = 3√3/6√3 = 1/2
i.e. AB/RQ = BC/QP = CA/PR
So, Δ ABC ~ Δ RQP (SSS similarity)
Therefore, ∠ C = ∠ P (Corresponding angles of similar triangles)
But ∠ C = 180o – ∠ A – ∠ B (Angle sum property)
= 180o – 80o – 60o
= 40o
So, ∠ P = 40o
Example-6 :- In Figure, OA . OB = OC . OD. Show that ∠ A = ∠ C and ∠ B = ∠ D.
Given that : OA . OB = OC . OD
Prove that : ∠ A = ∠ C and ∠ B = ∠ D
Proof : We have OA . OB = OC . OD (Given)
So, OA/OC = OD/OB ......(i) (Alternate angles)
Also, we have ∠ AOD = ∠ COB .....(ii) (Vertically opposite angles)
Therefore, from (i) and (ii),
Δ AOD ~ Δ COB (SAS similarity criterion)
So, ∠ A = ∠ C and ∠ D = ∠ B (Corresponding angles of similar triangles)
Example-7 :- A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post.
In figure, DE is the shadow of the girl. Let DE be x metres.
Now, BD = 1.2 m × 4 = 4.8 m.
In Δ ABE and Δ CDE,
∠ B = ∠ D (Each is of 90o because lamp-post as well as the girl are standing vertical to the ground)
and ∠ E = ∠ E (Same angle)
So, Δ ABE ~ Δ CDE (AA similarity criterion)
Therefore, (4.8 + x)/x = 3.6/0.9
4.8 + x = 4x
4.8 = 4x - x
4.8 = 3x
4.8/3 = x
x = 1.6
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
Example-8 :- In Figure, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, prove that :
(i) Δ AMC ~ Δ PNR
(ii) CM/RN = AB/PQ
(iii) Δ CMB ~ Δ RNQ
Given that : CM and RN are respectively the medians of Δ ABC and Δ PQR. Δ ABC ~ Δ PQR
Prove that :(i) Δ AMC ~ Δ PNR
(ii) CM/RN = AB/PQ
(iii) Δ CMB ~ Δ RNQ
Proof : (i) We have Δ ABC ~ Δ PRQ (Given)
So, AB/PQ = BC/QR = CA/RP ......(i)
and ∠ A = ∠ P, ∠ B = ∠ Q and ∠ C = ∠ R .....(ii)
AB = 2 AM and PQ = 2 PN (As CM and RN are medians)
So, from (i), 2AM/2PN = CA/RP
i.e., AM/PN = CA/RP .......(iii)
Also, ∠ MAC = ∠ NPR [From (ii)] ......(iv)
So, from (iii) and (iv),
Δ AMC ~ Δ PNR .......(v) (SAS similarity)
(ii) From (v) CM/RN = CA/RP ....(vi)
CA/RP = AB/PQ [from (i)] ....(vii)
Therefore, CM/RN = AB/PQ [from (vi) and (vii)] .....(viii)
(iii) Again, AB/PQ = BC/QR [frrom (i)]
Therefore, CM/RN = BC/QR [from (viii)] .....(ix)
Also, CM/RN = AB/PQ = 2BM/2QN
CM/RN = BM/QN .......(x)
CM/RN = BC/QR = BM/QN [from (ix) and (x)]
Therefore, Δ CMB ~ Δ RNQ (SSS similarity)
Example-9 :- In Figure, the line segment XY is parallel to side AC of Δ ABC and it divides the triangle into two parts of equal areas. Find the ratio AX/AB.
Given that The line segment XY is parallel to side AC of Δ ABC.
i.e., XY || AC
So, ∠ BXY = ∠ A and ∠ BYX = ∠ C (Corresponding angles)
Therefore, Δ ABC ~ Δ XBY (By AA similarity Rule)
So, ar(ABC)/ar(XBY) = (AB/XB)2....(i) (Therorem-6)
Also given It divides the triangle into two parts of equal areas.
i.e., ar (ABC) = 2 ar (XBY)
So, ar(ABC)/ar(XBY) = 2/1 .........(ii)
Therefore, from (i) and (ii),
(AB/XB)2 = 2/1
AB/XB = √2/1
XB/AB = 1/√2
1 - XB/AB = 1 - 1/√2
(AB - XB)/AB = (√2 - 1)/√2
AX/AB = (2 - √2)/2
Example-10 :- In Figure, ∠ ACB = 90o and CD ⊥ AB. Prove that BC2/AC2 = BD/AD.
Given that : ∠ ACB = 90o and CD ⊥ AB.
Prove that : BC2/AC2 = BD/AD
Proof : We have ∠ ACB = 90o and CD ⊥ AB (Given)
Δ ACD ~ Δ ABC (Theorem-7)
So, AC/AB = AD/AC
AC2 = AB . AD ......(i)
Similarly, Δ BCD ~ Δ BAC (Theorem-7)
So, BC/BA = BD/BC
BC2 = BA . BD ......(ii)
Therefore, from (i) and (ii),
BC2/AC2 = (BA . BD)/(AB . AD) = BD/AD
Example-11 :- A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.
Solution :-Let AB be the ladder and CA be the wall with the window at A. Also, BC = 2.5 m and CA = 6 mFrom Pythagoras Theorem, we have: AB2 = BC2 + CA2 AB2 = (2.5)2 + (6)2 AB2 = 42.25 AB = 6.5 Thus, length of the ladder is 6.5 m.
Example-12 :- In Figure, if AD ⊥ BC, prove that AB2 + CD2 = BD2 + AC2.
Given that : AD ⊥ BC
Prove that : AB2 + CD2 = BD2 + AC2
Proof : In Δ ADC, we have AC2 = AD2 + CD2 ....(i) (By Pythagoras Theorem)
In Δ ADB, we have AB2 = AD2 + BD2 ....(ii)(By Pythagoras Theorem)
By Subtracting (i) from (ii), we get
AB2 – AC2 = BD2 – CD2
AB2 + CD2 = BD2 + AC2
Example-13 :- BL and CM are medians of a triangle ABC right angled at A. Prove that 4(BL2 + CM2) = 5BC2.
Given that : BL and CM are medians of the Δ ABC in which ∠ A = 90o
Prove that : 4(BL2 + CM2) = 5BC2
Proof : In Δ ABC, we have BC2 = AB2 + AC2 ....(i) (By Pythagoras Theorem)
In Δ ABL, we have BL2 = AL2 + AB2 (By Pythagoras Theorem)
BL2 = (AC/2)2 + AB2 (L is the mid point of AC)
BL2 = AC2/4 + AB2
4BL2 = AC2 + 4AB2 .......(ii)
In Δ CMA, we have CM2 = AC2 + AM2 (By Pythagoras Theorem)
CM2 = AC2 + (AB/2)2 (M is the mid point of AB)
CM2 = AC2 + AB2/4
4CM2 = 4AC2 + AB2 .......(iii)
By Adding (ii) and (iii), we get
4(BL2 + CM2) = 5(AC2 + AB2)
4(BL2 + CM2) = 5BC2 [From (i)]
Example-14 :- O is any point inside a rectangle ABCD (see Figure). Prove that OB2 + OD2 = OA2 + OC2.
Given that : O is any point inside a rectangle ABCD. Prove that : OB2 + OD2 = OA2 + OC2 Construction : Draw PQ || BCProof : Through O, draw PQ || BC so that P lies on AB and Q lies on DC. Now, PQ || BC Therefore, PQ ⊥ AB and PQ ⊥ DC(∠ B = 90o and ∠ C = 90o) So, ∠ BPQ = 90o and ∠ CQP = 90o Therefore, BPQC and APQD are both rectangles. Now, in Δ OPB, OB2 = BP2 + OP2 ...... (i) Similarly, in Δ OQD, OD2 = OQ2 + DQ2 ...... (ii) Similarly, in Δ OQC, OC2 = OQ2 + CQ2 ...... (iii) Similarly, in Δ OAP, OA2 = AP2 + OP2 ...... (iv) By Adding (i) and (ii), OB2 + OD2 = BP2 + OP2 + OQ2 + DQ2 OB2 + OD2 = CQ2 + OP2 + OQ2 + AP2 (As BP = CQ and DQ = AP) OB2 + OD2 = CQ2 + OQ2 + OP2 + AP2 OB2 + OD2 = OC2 + OA2 [From (iii) and (iv)]
