Answer:
See Explanation
Step-by-step explanation:
[tex] m\angle ADB = m\angle CDB... (given) \\\\
\therefore m\angle ADM = m\angle CDM... (1)\\
(\because D-M-B) \\\\
In\: \triangle 's ADM \: \&\: CDM\\\\
\overline{AD} \cong\overline {CD} ... (given) \\\\
m\angle ADM = m\angle CDM.(From \:1)\\\\
\overline{DM} \cong\overline{DM} ... (given) \\\\
\therefore \triangle ADM \: \cong\: \triangle CDM\\.. (By \: SAS\: postulate) \\\\
\therefore \overline{AM} \cong\overline{CM}..(2)\\(by\: c. s. c. t.) \\\\
m\angle AMD = m\angle CMD..(3)\\(by\: c. a. c. t.)\\\\
\because m\angle AMD + m\angle CMD= 180\degree ..(4)\\
(Linear\: pair\: \angle 's)\\\\
\therefore m\angle AMD + m\angle AMD= 180\degree \\..(From \: 3 \: \& \: 4)\\\\
\therefore 2m\angle AMD = 180\degree\\\\
\therefore m\angle AMD = \frac{180\degree}{2}\\\\
\therefore m\angle AMD =90\degree \\\\
\red{\implies \overline{MD} \perp\overline{AM}} \\\\
\implies m\angle AMB=m\angle CMB= m\angle CMD = 90\degree.. (5)\\\\
In\: \triangle 's ABM \: \&\: CBM\\\\
\overline{AM} \cong\overline{CM}\\.(From \: 2)\\\\
m\angle AMB=m\angle CMB\\..(each\: 90\degree) \\\\
\overline{BM} \cong\overline{BM} ... (common) \\\\
\therefore \triangle ABM \: \cong\: \triangle CBM\\.. (By \: SAS\: postulate) \\\\
\therefore m\angle BAM = m\angle BCM\\(by\: cact) \\\\
\purple {\implies m\angle BAC = m\angle BCA} \\(\because A-M-C) [/tex]
