Question

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Answer 1

$\bigstar \frak{Required \: Solution }$

Question 1 :-

• ABCD is a parallelogram

Reason:-

Here:

▪︎AB//DC

▪︎AD//BC

.°. the given shape is parallogram

• AB//DC

Reason :-

opposite sides of parallelogram are parallel to each other.

.°. AB//DC

• Alternative interior angles are equal

Statement:-

alternate interior angles

• opposite sides of parallelogram are congruent

Statement:-

AB = DC, AD = BC

• DGC ≅ BEA

Reason:-

ASA postulate

• CPCT

Statement:

AE = EC

BE = ED

------------------------------------------------------------

Question 2:-

• Given

Statement:

BIKE is a rectangle

• BE = IK

Reason:-

Opposite sides of rectangle are equal

• All right angles are equal

Statement:

∠A = ∠B = ∠C = ∠D

• SAS postulate

Statement :

△EBI = △EKI

• CPCT

Statement :

∠BIE = ∠KEI

Answer 2

Answer:

$\bigstar \frak{Required \: Solution }★RequiredSolution</p><p></p><p>Question 1 :-</p><p></p><p>ABCD is a parallelogram</p><p></p><p>Reason:-</p><p></p><p>Here:</p><p></p><p>▪︎AB//DC</p><p></p><p>▪︎AD//BC</p><p></p><p>.°. the given shape is parallogram</p><p></p><p>AB//DC</p><p></p><p>Reason :-</p><p></p><p>opposite sides of parallelogram are parallel to each other.</p><p></p><p>.°. AB//DC</p><p></p><p>Alternative interior angles are equal</p><p></p><p>Statement:-</p><p></p><p>alternate interior angles</p><p></p><p>opposite sides of parallelogram are congruent</p><p></p><p>Statement:-</p><p></p><p>AB = DC, AD = BC</p><p></p><p>DGC ≅ BEA</p><p></p><p>Reason:-</p><p></p><p>ASA postulate</p><p></p><p>CPCT</p><p></p><p>Statement:</p><p></p><p>AE = EC</p><p></p><p>BE = ED</p><p></p><p>------------------------------------------------------------</p><p></p><p>Question 2:-</p><p></p><p>Given</p><p></p><p>Statement:</p><p></p><p>BIKE is a rectangle</p><p></p><p>BE = IK</p><p></p><p>Reason:-</p><p></p><p>Opposite sides of rectangle are equal</p><p></p><p>All right angles are equal</p><p></p><p>Statement:</p><p></p><p>∠A = ∠B = ∠C = ∠D</p><p></p><p>SAS postulate</p><p></p><p>Statement :</p><p></p><p>△EBI = △EKI</p><p></p><p>CPCT</p><p></p><p>Statement :</p><p></p><p>∠BIE = ∠KEI</p><p></p><p>$