Answers to the eighth grade mathematics textbook published by Beijing Normal University (1)
Exercise on page 20
1. Solution: (1) Pseudo-proposition. As shown in figure 1-2-34,
In Rt△ABC and Rt△A'B'C? Yes,? A=？ A'=90？ ,
? B=？ C=45？ =? b？ =? c？ ，AB= AC？ B? =A'C？ , then Rt△ABC and Rt△A'B'C? Inequality,
(2) true proposition,
Known: as shown in figure 1-2-35,? C=？ c？ =90? ,? A=？ Answer? ，AB=A'B '。
Proof: Rt△A BC≌Rt△A'B'C? .
Prove:
∵? C=？ c？ = 90? ,? A=？ Answer? ，AB=A'B '，
? Rt△ABC≌Rt△A'B'C？ (AAS)。
(3) true proposition,
Known: as shown in figure 1-2-35,? C=？ c？ =90? ，AC=A'C '，BC=B'C '。
Proof: Rt△ABC≌Rt△A'B'C? .
Prove:
AC = A 'C？ ,? C=？ c？ =90? ，BC=B？ c？ ,
? Rt△ABC≌Rt△A？ BC? (SAS)。
(4) True proposition
Known: as shown in figure 1-2-36,? C=？ c？ =90? ,
AC=A？ c？ , midline AD=A'D'.
Verification: Rt△ABC≌RtAA'B'C? .
Prove:
∵? C=？ c？ =90? ，AD=AD？ ，AC=A'C？ ,
? Rt△ACD≌Rt△A'C'D'(HL)。
? DC = Washington? .
∫ BC =2D, BC =2D,
? BC = BC?
? Rt△ABC≌Rt△A'B'C(SAS)。
2. Solution: The same reason:
∫AB = AC = 12m。
? A triangle consisting of three points A, B and C is an isosceles triangle.
∵ and ao BC.
? AO is the median line on the base BC of the isosceles triangle △ABC,
? BO=CO，
? Twenty stakes are equidistant from the bottom of the Xuan.
Answers to the eighth grade mathematics textbook published by Beijing Normal University (II)
Exercise 1.6
1. Proof:
∫D is the midpoint of BC,
? BD=CD。
At Rt△BDF and Rt△CDE,
? Rt△BDF≌Rt△CDE(HL)。
B=？ C (the corresponding sides of congruent triangles are equal),
? AB=AC (equilateral),
? △ABC is an isosceles triangle.
2. Prove:
∵ Germany? AC，BF？ Communication,
DEC=？ BFA=90？ .
At Rt△ABF and Rt△CDE,
? Rt△ABF≌Rt△CDE(HL)。
? AF=CE，？ A=？ C (congruent triangles has equal corresponding edges and equal corresponding angles).
? AB//CD，AF-EF=CE-RF，
? AE=CF。
3. Prove:
∵MP？ OA，NP？ OB，
PMO=？ PNO=90？ .
OM = on, OP=OP,
? Rt△POM≌Rt△PON(HL)。
AOP=？ BOP, that is, OP split equally? AOP。
4. Solution: (1) Pseudo-proposition. When two right angles of a right triangle are equal to a right side and a hypotenuse of another right triangle, two right triangles are not equal.
(2) false proposition. When an acute angle and a right-angled side of a right-angled triangle are equal to an acute angle and a hypotenuse of another right-angled triangle, two right-angled triangles are not equal.
5.( 1) solution: edge: DB=DA, BE = AE horn:? B=？ Bad =30? ,? ADE=？ BDE=60？ ,? Bed =? AED=90？ .
(2) prove that:
∵? C=90？ ,? B=30？ ,
BAC=60？ .
∵? Bad =? B=30？ .
CAD=？ EAD=30？ .
Again? AED=？ C=90？ , and AD=AD,
? △ACD?△AED(AAS)。
The proof of this problem is not unique.
(3) no.
Answers to the second volume of the eighth grade mathematics textbook published by Beijing Normal University (3)
Page 23
Prove:
∫AB is the angular bisector of the line segment CD,
? ED=EC, FC=FD (the property theorem of the vertical line in the line segment).
ECD=？ EDC (equilateral angle),? FCD=？ Equal sides and angles.