WORKSHEET - 1
Q.1 Diagonal AC of a parallelogram ABCD bisects ∠A . Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Q.2 In ∆ABC and ∆DEF, AB = DE, AB || DE, BC – EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see attached figure). Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF
Q.3 ABCD is a trapezium in which AB || CD and AD = BC. Show that
(i )∠A=∠B
(ii )∠C=∠D
(iii) ∆ABC ≅ ∆BAD
(iv) diagonal AC = diagonal BD
Answers
Answer:
Q no 1= ABCD is a rhombus
Step-by-step explanation:
Q no1=
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[ Alternate angles ] --- ( 2 )
AD∥BC and AAC as traversal,
∴ ∠DAC=∠BCA [ Alternate angles ] --- ( 3 )
From ( 1 ), ( 2 ) and ( 3 )
∠DAC=∠BAC=∠DCA=∠BCA
∴ ∠DCA=∠BCA
Hence, AC bisects ∠C.
(ii) In △ABC,
⇒ ∠BAC=∠BCA [ Proved in above ]
⇒ BC=AB [ Sides opposite to equal angles are equal ] --- ( 1 )
⇒ Also, AB=CD and AD=BC [ Opposite sides of parallelogram are equal ] ---- ( 2 )
From ( 1 ) and ( 2 ),
⇒ AB=BC=CD=DA
Q no2=
(i) Consider the quadrilateral ABED
We have , AB=DE and AB∥DE
One pair of opposite sides are equal and parallel. Therefore
ABED is a parallelogram.
(ii) In quadrilateral BEFC , we have
BC=EF and BC∥EF. One pair of opposite sides are equal and parallel.therefore ,BEFC is a parallelogram.
(iii) AD=BE and AD∥BE ∣ As ABED is a ||gm ... (1)
and CF=BE and CF∥BE ∣ As BEFC is a ||gm ... (2)
From (1) and (2), it can be inferred
AD=CF and AD∥CF
(iv) AD=CF and AD∥CF
One pair of opposite sides are equal and parallel
⇒ ACFD is a parallelogram.
(v) Since ACFD is parallelogram.
AC=DF ∣ As Opposite sides of a|| gm ACFD
(vi) In triangles ABC and DEF, we have
AB=DE ∣ (opposite sides of ABED
BC=EF ∣ (Opposite sides of BEFC
and CA=FD ∣ Opposite. sides of ACFD
Using SSS criterion of congruence,
△ABC≅△DEF
Q no 3 =
Given ABCD is trapezium where AD=BC.
(i) To prove: ∠A=∠B
we can see that AECD is a parallelogram, so sum of adjacent angles =180
o
→∠A+∠E=180
o
→∠A+x=180
o
→∠A=180
o
−x=∠B
Hence proved.
(ii) To prove: ∠C=∠D
sum of adjacent angles in parallelogram is π, so
→∠D∠C+180
o
−2x=180
o
→∠C+∠D=2x
Now
→∠B+∠C=180
o
→180
o
−x+∠C=180
o
=0 ∠C=x, so
∠D=x
And,
∠C=∠D
Hence proved.
(iii) ΔABC=ΔBAD
→ side AB is common.
→AD=BC (given)
so the angle including both the sides is also same,
∠A=∠B. So
ΔABC=ΔBAD (By SAS congruent Rule)
Hence proved.
(iv) As ΔABC=ΔBAD
The third side of both triangles i.e. diagonals are equal AC=BD