30. triangle ABC is an Isosceles …
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Amar Kumar 8 years, 6 months ago
{tex}\eqalign{ & Given\,that\,:ABC\,is\,a\,triangle\,\,in\,which\, \cr & AB = AC \cr & \angle ABC = \angle ACB\,\,\,\,\,\,\,(1) \cr & AB = AD \cr & join\,\,CD \cr & In\,\,\Delta ACD \cr & AC = AD \cr & \angle ACD = \angle ADC\,\,\,\,(2) \cr & In\,\,\Delta BCD\,from\,eq.\,\,(1)\,\,and\,\,(2) \cr} {/tex}
{tex}\eqalign{ & \angle DBC = \,\angle BDC \cr & \angle DBC + \angle BDC + \angle BCD = 180^\circ \cr & 2\angle BDC + \angle BCD = 180^\circ \cr & 2\angle BDC = \angle BCD \cr & 4\angle BDC = 180^\circ \cr & \angle BDC = 45^\circ \cr & \angle DBC = 45^\circ \cr & So\,\,\angle BCD = 90^\circ \cr}{/tex}
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