Question

In the given figure, ACDH is a parallelogram, EFGH is a square and the straight line BEG is parallel to CD. Find $\angle CDE$

• A. ${45}^0$
• B. ${125}^0$
• C. ${135}^0$
• D. ${185}^0$

Answer 1

The correct option is C ${135}^0$

Line BEG bisects $\angle HEF.$

So, $\angle HEG={45}^{\circ }$

Now, $\angle BEH={180}^{\circ }-\angle HEG$

$\Rightarrow \angle BEH={180}^{\circ }-{45}^{\circ }={135}^{\circ }$

$BE\parallel CD$

So, $\angle BEH=\angle CDE={135}^{\circ }$ (Pair of corresponding angles)

Hence, C is the correct option.