<u>Solving Q1: Given BE = ED and AE = EC.</u>
Considering triangles ΔAED and ΔCEB;
1. AE = EC (given)
2. ∠AED = ∠CEB (vertical angles)
3. ED = EB (given)
ΔAED ≡ ΔCEB (Side-Angle-Side congruency of triangles)
∠DAE = ∠BCE and AD = BC (using CPCTC)
⇒ AD║BC (Converse of Alternate Interior angles theorem)
Similarly consider triangles ΔABE and ΔCDE;
1. AE = CE (given)
2. ∠BEA = ∠DEC (vertical angles)
3. BE = DE (given)
ΔABE ≡ ΔCDE (Side-Angle-Side congruency of triangles)
∠ABE = ∠CDE and AB = CD (using CPCTC)
⇒ AB║CD (Converse of Alternate Interior angles theorem)
Since we have AB║CD, AB = CD and AD║BC, AD = BC.
Therefore, quadrilateral ABCD is a parallelogram.
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<u>Solving Q2: The parallelogram has the angle measures shown in the diagram.</u>
It is clearly visible that both the triangles are Isosceles triangles, so opposite sides in each triangle are equal.
Consider two triangles given in the problem, we have two sets of congruent angles and one included side is common in both triangles.
Using Angle-Side-Angle congruence of triangles, both the triangles would be congruent too.
Using CPCTC, we can say opposite sides of quadrilateral would be congruent.
Therefore, all sides in given quadrilateral are equal.
Hence, Given quadrilateral is a Rhombus.
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