guys please help answer fast with full process please give answer fast tommorow is my exam
Answers
Step-by-step explanation:
In △EDC and △EBA, we have
∠1=∠2 [Alternate angles]
∠3=∠4 [Alternate angles] and,
∠CED=∠AEB [Vertically opposite angles]
∴ △EDC∼△EBA
⇒
EB
ED
=
EA
EC
⇒
EC
ED
=
EA
EB
.............(I)
It is given that △AED∼△BEC
∴
EC
ED
=
EB
EA
=
BC
AD
From (i) and (ii), we get
EA
EB
=
EB
EA
⇒ (EB)
2
=(EA)
2
⇒ EB=EA
Substituting EB=EA in (ii), we get
EA
EA
=
BC
AD
⇒
BC
AD
=1
⇒ AD=BC [Hence proved]
Answer:
(i) DE║AB (∵DC║AB)
DA║EB (given)
⇒ABED is a parallelogram
(ii) ∠C=∠D (∵AD=BC) [1]
∠A=∠B (∵AD=BC) [2]
∠A+∠B+∠C+∠D = 360°(angle sum property of a quadrilateral)
∠A+∠A+∠C+∠C = 360° (by [1] and [2])
2(∠A+∠C) = 360°
∠A+∠C = 180°
Similarly, ∠B+∠D = 180°
-------------------------------------------------------------------------