问题1.在四边形ACBD中， AC = AD，AB将等分∠A(见图7.16)。证明∆ ABC≅∆ ABD您对BC和BD有什么看法?
解决方案：
Given that AC and AD are equal
i.e. AC = AD and the line AB bisects ∠A.
Considering the two triangles ΔABC and ΔABD,
Where,
AC = AD { As given}………………………………………… (i)
∠CAB = ∠DAB ( As AB bisects of ∠A)……………. (ii)
AB { Common side of both the triangle} …….. …(iii)
From above three equation both the triangle satisfies “SAS” congruency criterion
So, ΔABC ≅ ΔABD.
Also,
BC and BD will be of equal lengths as they are corresponding parts of congruent triangles(CPCT).
So BC = BD.
问题2。ABCD是一个四边形，其中AD = BC，DAB = CBA(见图7.17)。证明
(i)∆ ABD≅∆ BAC
(ii)BD = AC
(iii)∠ABD =∠BAC。
解决方案：
(i) Given that AD = BC,
And ∠ DAB = ∠ CBA.
Considering two triangles ΔABD and ΔBAC.
Where,
AD = BC { As given }………………………………………….. (i)
∠ DAB = ∠ CBA { As given also}……………………….. (ii)
AB {Common side of both the triangle)…………. (iii)
From above three equation two triangles ABD and BAC satisfies “SAS” congruency criterion
So, ΔABD ≅ ΔBAC
(ii) Also,
BD and AC will be equal as they are corresponding parts of congruent triangles(CPCT).
So BD = AC
(iii) Similarly,
∠ABD and ∠BAC will be equal as they are corresponding parts of congruent triangles(CPCT).
So,
∠ABD = ∠BAC.
问题3. AD和BC垂直于线段AB(见图7.18)。证明CD对分AB。
解决方案：
Given that AD and BC are two equal perpendiculars to a line segment AB
Considering two triangles ΔAOD and ΔBOC
Where,
∠ AOD = ∠ BOC {Vertically opposite angles}………………. (i)
∠ OAD = ∠ OBC {Given that they are perpendiculars}…. (ii)
AD = BC {As given}………………………………………………… (iii)
From above three equation both the triangle satisfies “AAS” congruency criterion
So, ΔAOD ≅ ΔBOC
AO and BO will be equal as they are corresponding parts of congruent triangles(CPCT).
So, AO = BO
Hence, CD bisects AB at O.
问题4. l和m是两条平行线，与另一对平行线p和q相交(见图7.19)。证明∆ ABC≅∆ CDA。
解决方案：
Given that l and m are two parallel lines p and q are another pair of parallel lines
Considering two triangles ΔABC and ΔCDA
Where,
∠ BCA = ∠DAC {Alternate interior angles}…. (i)
∠ BAC = ∠ DCA {Alternate interior angles}…. (ii)
AC {Common side of two triangles}………….(iii)
From above three equation both the triangle satisfies “ASA” congruency criterion
So, ΔABC ≅ ΔCDA
问题5.线l是角度∠A的等分线，而B是l上的任意点。 BP和BQ从B到∠A的臂垂直(见图7.20)。显示：
(i)∆ APB≅∆ AQB
(ii)BP = BQ或B与∠A的臂等距。
解决方案：
Given that, Line l is the bisector of an angle ∠ A and B
BP and BQ are perpendiculars from angle A.
Considering two triangles ΔAPB and ΔAQB
Where,
∠ APB = ∠ AQB { Two right angles as given }…… (i)
∠BAP = ∠BAQ (As line l bisects angle A }……… (ii)
AB { Common sides of both the triangle }……… (iii)
From above three equation both the triangle satisfies “AAS” congruency criterion
So, ΔAPB≅ ΔAQB.
(ii) Also we can say BP and BQ are equal as they are corresponding parts of congruent triangles(CPCT).
So, BP = BQ
问题6.在图7.21中，AC = AE，AB = AD，∠BAD =∠EAC。证明BC = DE。
解决方案：
Given that AC = AE, AB = AD
And ∠BAD = ∠EAC
As given that ∠BAD = ∠EAC
Adding ∠DAC on both the sides
We get,
∠BAD + ∠DAC = ∠EAC + ∠ DAC
∠BAC = ∠EAD
Considering two triangles ΔABC and ΔADE
Where,
AC = AE { As given }…………………… (i)
∠BAC = ∠EAD { Hence proven }…….. (ii)
AB = AD {As also given }……………….. (iii)
From above three equation both the triangle satisfies “SAS” congruency criterion
So, ΔABC ≅ ΔADE
(ii) Also we can say BC and DE are equal as they are corresponding parts of congruent triangles(CPCT).
So that BC = DE.
问题7. AB是线段，P是其中点。 D和E是AB的同一侧上的点，因此∠BAD =∠ABE和∠EPA =∠DPB(见图7.22)。显示：
(i)∆ DAP≅∆ EBP
(ii)AD = BE
解决方案：
Given that P is the mid-point of line AB, So AP = BP
Also, ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB
Now adding ∠DPE on both the sides of two equal angle ∠ EPA = ∠ DPB
∠ EPA + ∠ DPE = ∠ DPB + ∠DPE
Which implies that two angles ∠ DPA = ∠ EPB
Considering two triangles ∆ DAP and ∆ EBP
∠ DPA = ∠ EPB { Hence proven }…… (i)
AP = BP { Hence Given }……………… (ii)
∠ BAD = ∠ ABE { As given }…………..(iii)
From above three equation both the triangle satisfies “ASA” congruency criterion
So, ΔDAP ≅ ΔEBP
(ii) Also we can say AD and BE are equal as they are corresponding parts of congruent triangles(CPCT).
So that, AD = BE
问题8。在直角三角形ABC中，与C成直角，M是斜边AB的中点。 C连接到M并产生到D点，使得DM = CM。 D点与B点相连(见图7.23)。显示：
(i)∆ AMC≅∆ BMD
(ii)∠DBC是直角。
(iii)∆ DBC≅∆ ACB
(iv)CM = 1/2 AB
解决方案：
Given that M is the mid-point of AB
So AM = BM
∠ ACB = 90°
and DM = CM
(i) Considering two triangles ΔAMC and ΔBMD:
AM = BM { As given }………………………………………. (i)
∠ CMA = ∠ DMB { Vertically opposite angles }…. (ii)
CM = DM { As given also}……………………………….. (iii)
From above three equation both the triangle satisfies “SAS” congruency criterion
So, ΔAMC ≅ ΔBMD
(ii) From above congruency we can say
∠ ACD = ∠ BDC
Also alternate interior angles of two parallel lines AC and DB.
Since sum of two co-interiors angles results to 180°.
So, ∠ ACB + ∠ DBC = 180°
∠ DBC = 180° – ∠ ACB
∠ DBC = 90° { As ∠ACB =90° }
(iii) In ΔDBC and ΔACB,
BC { Common side of both the triangle }……. (i)
∠ ACB = ∠ DBC { As both are right angles }….(ii)
DB = AC (by CPCT)………………………………….. (iii)
From above three equation both the triangle satisfies “SAS” congruency criterion
So, ΔDBC ≅ ΔACB
(iv) As M is the mid point so we can say
DM = CM = AM = BM
Also we can say that AB = CD ( By CPCT )
As M is the mid point of CD we can write
CM + DM = AB
Hence, CM + CM = AB (As DM = CM )
Hence, CM = (½) AB