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)]