Question

prove that it's important and get 15 points​

Answer 1

$\bf{\Huge{\boxed{\rm{\blue{ANSWER\::}}}}}$

$\bf{Given}\begin{cases}\rm{AB\:=\:AC}\\\\ \rm{AD\:=\:AB}\end{cases}}$

$\bf{\Large{\underline{\bf{To\:prove\::}}}}$

$\rm{\underbrace{\rm{\angle BCD\:=\:90 \degree}}}$

$\bf{\Large{\underline{\tt{\green{Proof\::}}}}}$

$\bf{\large{In\:\triangle ABC,}}$

$\rm{AB\:=\:AC\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:[Given]}$

$\rm{\angle ACB\:=\:\angle ABC\:\:\:\:\:\:\:\:\:\:\:\:\:\:[Angle\:opposite\:\&\:sides\:are\:equal]}$

&

$\bf{\large{In\:\triangle ACD,}}$

$\rm{AC\:=\:AD\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:[Given]}$

$\rm{\angle ADC\:=\:\angle ACD\:\:\:\:\:\:\:\:\:\:\:\:\:\:[Angle\:opposite\:\&\:sides\:are\:equal]}$

___________________________________________________

$\bf{\large{In\:\triangle BCD,}}$

$\rm{\angle ABC\:+\:\angle BCD\:+\:\angle BCD\:=\:180\:\:\:\:\:\:\:\:\:\:\:[Property\:of\:\triangle]}$

$\rm{\angle ACB\:+\:\angle BCD\:+\:\angle ACD\:=\:180\:\:\:\:\:\:\:\:\:\:\:[From\:above\:both\:angle]}$

$\rm{(\angle ACB\:+\:\angle ACD)\:+\:\angle BCD\:=\:180}$

$\rm{\angle BCD\:+\:\angle BCD\:=\:180}$

$\rm{2\angle BCD\:=\:180}$

$\rm{\angle BCD\:=\:\cancel{\frac{180}{2} }}$

$\rm{\pink{\angle BCD\:=\:90 \degree}}$

Hence,

∠BCD is a right angled Δ.                                [Proved]

Answer 2

Answer:

${\boxed{\bf\green{Answer\::}}}$

Step-by-step explanation:

$\bf{Given}\begin{cases}\text{AB = AC}\\\text{AD = AB}\\\sf{\angle BCD\: =90\degree \: \: [To\:prove]}\end{cases}$

$\tt In\: \: \: \triangle \: ABC, \\\sf AB = AC \\ \therefore\tt \angle ACB =\angle ABC$

___________________________

$\because\tt AD = AB \\\therefore\tt AC = AD \\\therefore\sf \angle ADC =\angle ACD$

$\sf We\:know\:that, \\ \sf Sum\:of\:3\:angles\:of\:triangle=180\degree$

____________________________

$\tt In\: \: \: \triangle BCD, \\\rightarrow\tt \angle DBC + \angle BCD +\angle BDC = 180\degree \\\rightarrow\tt \angle ABC + \angle BCD +\angle ADC =180\degree\\\rightarrow\tt \angle ACB +\angle BCD + \angle ACD =180\degree \\\rightarrow\tt (\angle ACB +\angle ACD) + \angle BCD =180\degree \\\rightarrow\tt \angle BCD +\angle BCD = 180\degree \\\rightarrow\tt 2(\angle BCD) =180\degree \\\rightarrow\tt \angle BCD =\frac{180\degree}{2}\\\rightarrow\tt \angle BCD =90\degree$

$\therefore\bold{\underline{\angle BCD \:is \:a\:right\:angle \: \: \: \: [Proved]}}$