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Question 11
Question: Points A, B are on the same side of line l. AD ^ l and BE ^ l in D and E respectively. C is the mid-point of AB. Prove that CD=CE.
Answer: Given:
D ^ l and BE ^ l.
C is the mid-point of AB.
To prove:
CD=CE.
Construction:
Draw CM ^ l.
Proof:
AD, CM and BE are ( 
AD, CM and BE are Parallel to each other perpendicular to l)
AC=CB ( 
C is the mid-point of AB)
Intercepts made by parallel lines AD,CM and BE on the other transversal l are also equal.
i.e., DM=ME
Consider the triangles, CMD and CME
CM=CM ( common side)
( 
each 900)
DM=ME ( proved above)
(SAS congruency condition)
CD=CE (corresponding parts of corresponding triangles)
Hence CD=DE
Question 12
Question: E, F are respectively the mid-points of non-parallel sides AD, BC of a trapezium ABCD. Prove that,
a) EF||AB
Answer: Given:
E and F are mid-points of AD and BC.
To prove:
a) EF||AB
Proof:
a)
E and F are mid-points of AD and BC. (given)
DE=AE and CF=BF ( ABCD is a trapezium)
AB||CD
Equal intercepts are made on the transversal AD and BC.
EF||AB
Hence (a) is proved.
b)
In D ADB, E is the mid-point of AD . (given)
EP||AB ( proved)
----(i)
In D BCD, F is the mid-point of BC . (given)
PF||DC
----(ii)
By adding (i) and (ii), we get
Hence (b) is proved.
Question 13
Question: In D ABC, AD is the median through A and E is the mid-point
Answer: Given:
AD is the median through A and E is the mid-point of AD.
To prove:
Construction:
Through D, draw DK||BF.
Proof:
Consider D ADK,
( given)
( by construction)
(
line drawn through the mid- point of one side of a triangle, parallel to another side,
bisects the third side.)
Consider D BCF,
( 
AD is the median)
DK||BF ( by construction)
(
line drawn through the mid-point of one side of a triangle parallel to another side, bisects the third side.)
From (i) and (ii),
(using (iii))
Question 14
Question: M and N divide AB of D ABC into three equal parts. Line segment MP and NQ are both parallel to BC and meet AC in P and Q respectively. Prove that P, Q divide AC into three equal parts.
Answer: Given:
In DABC,
AM=MN=NB
MP||BC and NQ||BC
To prove:
AP=PQ=QC
Construction:
Draw DE || BC through A.
Proof:
AM=MN=NB (given)
and MP||BC; NQ||BC; DE||BC ( given)
i.e., AM, MN and NB are equal intercepts made on transversal AB .
AC is a transversal, intercepts made on AC are AP, PQ and QC.
AP=PQ=QC
Hence the proof.
Question 15
Question: In the adjoining figure, AD||BE||CF and B is the mid-point of AC. CD intersects BE at X. Show that
i) DE = EF
iv) AD + CF = 2BE
Answer: Given:
AD||BE||CF
AB=BC
CD intersects BE at X.
To prove:
i) DE=EF
iv) AD+CF=2BE
Proof:
AD||BE||CF ( given)
DF is a transversal. ( given)
DF makes equal intercepts.
Since AC makes equal intercepts. ( theorem)
DE=EF ---(i) ( proved )
B is the mid-point of AC.
X is the mid-point of CD. ( BX||AD)
E is the mid-point of DF . ( proved)
X is the mid-point of BC. ( proved)
BE=BX+EX
2BE=AD+CF ---(iv) ( proved)
