Hey, can you solve this??

Dear Student,
Consider this figure for part (a)

 
$$\begin{gathered} Consider ∆AXM\hspace{0.17em}and∆CYM \\ AM=MC (Since M is the midpoint of AC) \\ \angle YCM=\angle XAM (Alternate interior angles) \\ AX=CY \left(Given\right) \\ \Rightarrow ∆AXM\cong ∆CYM (By SAS\hspace{0.17em}criteria) \\ Hence Proved \left(a\right) \end{gathered}$$
(b)Consider the figure for part (b)

$$\begin{gathered} Since ∆AXM\cong CYM \\ \Rightarrow \angle 1=\angle 2 (By CPCT) \\ \angle 1+\angle 3=180° (Supplementary angles) \\ \Rightarrow \angle 2+\angle 3=180° ( \angle 1=\angle 2) \\ So, XY\hspace{0.17em}is a straight line (because angle at the centre is 180, so XMY\hspace{0.17em}doent bend anywhere) \\ Hence Proved \left(b\right) \end{gathered}$$
Regards