3. ABCD is trapezium in which AB || CD. IF AD = BC, show that angleA = angleB and
angleC = angleD.
Answers
Answer:
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=
−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
Answer:
Answer
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
