Question

Write the complete proof in your paper homework and for online (only) complete the probing statement (if any) that is a part of your proof or related to it.

Answer 1

Answer:

See Explanation

Step-by-step explanation:

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