Question

In $\square$ABCD , side BC < side AD (Figure 5.32) side BC || side AD and if side BA $\cong$ side CD then prove that $\angle$ABC$\cong$ $\angle$DCB.

Answer 1

Construction: Draw a line CE || AB
CE || AB (By construction)
AE || BC (Give)
So, ABCE is a parallelogram.
Since, AB || CE
⇒ ∠BAE = ∠CED = ∠x (Corresponding angles) .....(1)
Now, BA $\cong$ CE (ABCE is a parallelogram)
and BA $\cong$ CD (Given)
So, CE $\cong$ CD
⇒ ∠CED = ∠CDE = ∠x (Isosceles triangle) .....(2)
From (1) and (2) we have
∠BAE = ∠CDE = ∠x .....(3)
∠CEA = 180º − ∠CED = 180º − ∠x
In a parallelogram, the opposite angles are equal.
So, ∠ABC = ∠CEA = 180º − ∠x .....(4)
Similarly, ∠BAE = ∠BCE = ∠x
In ∆CED,
$$\begin{gathered} \angle \mathrm{ECD}+\angle \mathrm{CED}+\angle \mathrm{EDC}=180° \\ \Rightarrow \angle \mathrm{ECD}+x+x=180º \\ \Rightarrow \angle \mathrm{ECD}=180º-2x \end{gathered}$$
Now ∠DCB = ∠BCE + ∠ECD = x + 180º − 2x = 180º − x
Thus, $\angle$ABC$\cong$ $\angle$DCB