问题1.说明图中的哪些三角形对相似。写下您用来回答问题的相似性准则，并以符号形式写下成对的相似三角形：
解决方案：
(i) We have, in ΔABC and ΔPQR,
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
By AAA similarity criterion, we have
∴ ΔABC ~ ΔPQR
(ii) We have, in ΔABC and ΔPQR,
Computing the ratios of corresponding sides,
AB/QR = 2.5/5 = 1/2
BC/RP = 2/4 = 1/2
CA/PQ = 3/6 = 1/2
AB/QR = BC/RP = CA/PQ
Now,
By SSS similarity criterion, we have
ΔABC ~ ΔQRP
(iii) We have, in ΔLMP and ΔDEF,
Now,
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
Computing the ratios of the corresponding sides,
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF = 2.7/5 = 27/50
Since, MP/DE = PL/DF ≠ LM/EF
Therefore, ΔLMP and ΔDEF are not similar.
(iv) In ΔMNL and ΔQPR, it is given,
Computing the ratios of the corresponding sides,
MN/QP = ML/QR = 1/2
And,
∠M = ∠Q = 70°
By SAS similarity criterion
ΔMNL ~ ΔQPR
(v) In ΔABC and ΔDEF, we have,
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Computing the ratios of the corresponding sides,
AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠ ∠F
Therefore, ΔABC and ΔDEF are not similar.
(vi) In ΔDEF,
By angle sum property of a triangle,
∠D + ∠E + ∠F = 180°
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
⇒ ∠F = 30°
Similarly, In ΔPQR,
By angle sum property of a triangle,
∠P + ∠Q + ∠R = 180
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
⇒ ∠P = 70°
Now, in ΔDEF and ΔPQR, we have
∠D = ∠P = 70° [equal corresponding angles]
∠F = ∠Q = 80°
∠F = ∠R = 30°
Therefore,
By AAA similarity criterion,
ΔDEF ~ ΔPQR
问题2。在图中，ΔODC等于1/4ΔOBA，BOC等于125°，CDO等于70°。查找∠DOC，∠DCO和∠OAB。
解决方案：
We know, DOB is a straight line, as given.
Therefore, ∠DOC + ∠COB = 180°
Also,
∠BOC = 125°
⇒ ∠DOC = 180° – 125°
= 55°
In ΔDOC
By angle sum property of a triangle,
∠DCO + ∠CDO + ∠DOC = 180°
∠CDO = 70°
⇒ ∠DCO + 70º + 55º = 180°
⇒ ∠DCO = 55°
Also, given,
ΔODC ∝ ¼ ΔOBA,
Therefore, ΔODC ~ ΔOBA.
Since, we know that the corresponding angles are equal in similar triangles,
Therefore, ∠OAB = ∠OCD
⇒ ∠ OAB = 55°
∠OAB = ∠OCD
∠OAB = 55°
问题3.梯形ABCD与AB ||的对角AC和BD DC在点O处彼此相交。使用两个三角形的相似性准则，表明AO / OC = OB / OD
解决方案：
In ΔDOC and ΔBOA,
Since, AB || CD, therefore the alternate interior angles are equal,
That is, ∠CDO = ∠ABO ….(i)
And,
∠DCO = ∠BAO …(ii)
Also, vertically opposite angles are equal;
That is,∠DOC = ∠BOA…(iii)
Hence, by AAA similarity criterion,
ΔDOC ~ ΔBOA [From eq(i) , (ii) and (iii)]
Since the triangles are similar, therefore the corresponding sides are proportional.
DO/BO = OC/OA
⇒OA/OC = OB/OD
Hence, proved.
问题4。在图中，QR / QS = QT / PR，∠1=∠2。证明ΔPQS〜ΔTQR。
解决方案：
We have,
In ΔPQR,
∠PQR = ∠PRQ
That is, PQ = PR .…(i)
Given,
QR/QS = QT/PR
By using eq(i), we obtain,
QR/QS = QT/QP …….(ii)
In ΔPQS and ΔTQR,
QR/QS = QT/QP [By using eq(ii)]
∠Q = ∠Q
By SAS similarity criterion, therefore,
ΔPQS ~ ΔTQR
问题5. S和T是ΔPQR的PR和QR边的点，因此∠P=∠RTS。表明ΔRPQ〜ΔRTS。
解决方案：
We know,
S and T are the points on sides PR and QR of ΔPQR.
And, ∠P = ∠RTS.
In ΔRPQ and ΔRTS,
Given,
∠RTS = ∠QPS
∠R = ∠R (Common angle)
By AAA similarity criterion,
∴ ΔRPQ ~ ΔRTS
问题6。在图中，如果ΔABE≅ΔACD，则表明ΔADE〜ΔABC。
解决方案：
Given, ΔABE ≅ ΔACD.
By CPCT, we have
∴ AB = AC ……….(i)
And, AD = AE …………(ii)
In ΔADE and ΔABC,
On dividing the eq.(ii) by eq(i), we obtain,
AD/AB = AE/AC
Also, by common angle,
∠A = ∠A
By SAS similarity criterion, we have,
∴ ΔADE ~ ΔABC
问题7.在图中，ΔABC的高度AD和CE在点P处相交。
(i)ΔAEP〜ΔCDP
(ii)ΔABD〜ΔCBE
(iii)ΔAEP〜ΔADB
(iv)ΔPDC〜ΔBEC
解决方案：
Given, altitudes AD and CE of ΔABC intersect each other at the point P.
(i) In ΔAEP and ΔCDP,
Since, both of the angles are 90° each.
∠AEP = ∠CDP
Since,
∠APE = ∠CPD (Vertically opposite angles)
By AA similarity criterion, we have,
ΔAEP ~ ΔCDP
(ii) In ΔABD and ΔCBE,
Since, both of the angles are 90° each.
∠ADB = ∠CEB
∠ABD = ∠CBE (Common Angles)
By AA similarity criterion, we have,
ΔABD ~ ΔCBE
(iii) In ΔAEP and ΔADB,
Since, both of the angles are 90° each.
∠AEP = ∠ADB
∠PAE = ∠DAB (Common angles)
By AA similarity criterion, we have,
ΔAEP ~ ΔADB
(iv) In ΔPDC and ΔBEC,
Since, both of the angles are 90° each.
∠PDC = ∠BEC
∠PCD = ∠BCE (Common angles)
By AA similarity criterion, we have,
ΔPDC ~ ΔBEC
问题8. E是平行四边形ABCD产生的AD侧的点，BE在F处与CD相交。表明ΔABE〜ΔCFB。
解决方案：
Given,
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.
Now,
In ΔABE and ΔCFB,
Since, the opposite angles of a parallelogram are equal, we have,
∠A = ∠C
Also, since AE || BC
∠AEB = ∠CBF (Alternate interior angles)
By AA similarity criterion,
∴ ΔABE ~ ΔCFB
问题9。在图中，ABC和AMP是两个直角三角形，分别与B和M成直角，证明：
(i)ΔABC〜ΔAMP
(ii)CA / PA = BC / MP
解决方案：
Given,
ABC and AMP are two right triangles, right-angled at B and M respectively.
(i) In ΔABC and ΔAMP, we have,
∠CAB = ∠MAP (common angles)
Since, both of the angles are equivalent to 90°
∠ABC = ∠AMP
By AA similarity criterion, we have,
∴ ΔABC ~ ΔAMP
(ii) As, ΔABC ~ ΔAMP (AA similarity criterion) [proved in part (i)]
Therefore, we know,
If two triangles are similar then the corresponding sides are always equal,
Hence, CA/PA = BC/MP
问题10：CD和GH分别是∠ACB和∠EGF的等分线，使得D和H分别位于ΔABC和ΔEFG的AB和FE侧。如果ΔABC〜ΔFEG，则表明：
(i)CD / GH = AC / FG
(ii)ΔDCB〜ΔHGE
(iii)ΔDCA〜ΔHGF
解决方案：
Given,
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.
(i) We have,
ΔABC ~ ΔFEG.
Therefore,
∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
Now, since,
∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE
In ΔACD and ΔFGH,
∠A = ∠F
∠ACD = ∠FGH
By AA similarity criterion, we have,
∴ ΔACD ~ ΔFGH
⇒CD/GH = AC/FG
(ii) In ΔDCB and ΔHGE,
Proved in part (i)
∠DCB = ∠HGE
∠B = ∠E
By AA similarity criterion,
∴ ΔDCB ~ ΔHGE
(iii) In ΔDCA and ΔHGF,
Since, we have already proved,
∠ACD = ∠FGH
∠A = ∠F
By AA similarity criterion,
∴ ΔDCA ~ ΔHGF