Question

in triangle ABC, bisectors of angle A and angle C cross each other in point M. find angle ABC if it is half of AMC. (no shape given)
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Answer 1

Answer:

$\angle ABC=60$

Step-by-step explanation:

$[Kindly\ refer\ the\ image\ attachment.]\\We\ are\ given\ that,\\AM\ is\ the\ Angle\ Bisector\ of\ \angle BAC.\\Hence,\\\angle BAM= \angle MAC\\CM\ is\ the\ Angle\ Bisector\ of\ \angle BCA.\\Hence,\\\angle BCM= \angle MCA.\\Also,\\\angle AMC=2 \angle ABC$

$Now,\\As\ we\ can\ observe\ that,\\\angle BCM + \angle MCA= \angle BCA\\\angle MCA+ \angle MCA= \angle BCA\ [\angle BCM = \angle MCA]\\Hence,\\2 \angle MCA= \angle BCA\\Or, \\\angle MCA=\frac{\angle BCA}{2} \\\\Similarly,\\\angle BAM + \angle MAC= \angle BAC\\\angle MAC + \angle MAC= \angle BAC\\2 \angle MAC = \angle BAC\\Or,\\\angle MAC=\frac{\angle BAC}{2}$

$Through\ the\ Angle\ Sum\ Property\ of\ a\ Triangle,\ we\ know\ that:\\'The\ Sum\ of\ all\ interior\ angles\ of\ a\ triangle\ is\ 180.'\\Hence,\\In\ \triangle BAC,\\\angle ABC + \angle BCA + \angle CAB=180\\In\ \triangle MAC,\\\angle MAC+ \angle ACM + \angle AMC=180$

$Hence,\\As\ \angle AMC= 2 \angle ABC, \angle MCA=\frac{\angle BCA}{2}, \angle MAC=\frac{\angle BAC}{2},\\2 \angle ABC+\frac{\angle BCA}{2}+\frac{\angle BAC}{2}=180\\By\ resolving\ the\ denominators,\\\frac{4 \angle ABC+\angle BCA+\angle BAC}{2}=180\\\\By\ comparing\ the\ Sum\ of\ angles\ in\ both\ the\ triangles,\\We\ find\ that\ the\ RHS\ of\ both\ the\ equations\ are\ equal\ i.e.180,\\The\ LHS\ of\ the\ equations\ are\ equal\ too.$

$Hence,\\\angle ABC+ \angle BCA + \angle BAC=\frac{4 \angle ABC+\angle BCA+\angle BAC}{2}\\Hence,\\2(\angle ABC+ \angle BCA + \angle BAC)=4 \angle ABC+\angle BCA+\angle BAC\\Hence,\\2 \angle ABC+ 2 \angle BCA + 2 \angle BAC=4 \angle ABC+\angle BCA+\angle BAC\\Hence,\\2 \angle BCA + 2 \angle BAC-\angle BCA- \angle BAC=4 \angle ABC- 2 \angle ABC\\Hence,\\\angle BCA+\angle BAC= 2\angle ABC$

$Lets\ get\ back\ to\ the\ Angle\ Sum\ of\ \triangle ABC,\\\angle ABC + \angle BAC + \angle ACB=180\\Hence,\\As\ \angle BCA + \angle BAC= 2 \angle ABC,\\\angle ABC + 2 \angle ABC=180\\Hence,\\3 \angle ABC=180\\\angle ABC=\frac{180}{3}=60$