Question

If $\Delta ABC$ is an isosceles triangle in which AB = AC ,side BA is produced to D such that AD = AB. Show that $\angle BCD$ is a right angle.

Answer 1

Given that,

$AB=AC$

also,

$AD=AB$

i.e.,

$AC=AB=AD$

to prove:

$\angle BCD=90$

proof :

In $△ABC,$

$AB=AC$

[angle opposite to equal to sides are equal}

$\angle ACB=\angle ABC------\left(i\right)$

In $△ACD,$

$AC=AD$

[agle opposite to equal sides are equals]

$\angle ADC=\angle ACD---\left(ii\right)$

In$△BCD.,$

[apply sum property of $△$]

$=>\angle ABC+\angle BCD+\angle BDC=180$

from $\left(\right)$ and $\left(ii\right)$ we get,

$=>\angle ACB+\angle BCD+\angle ACD=180$

$=>(\angle ACB+\angle ACD)+\angle BCD=180$

$=>\left(\angle BCD\right)+\angle BCD=180$

$=>2\angle BCD={180}^{\circ }$

hence $\angle BCD={90}^{\circ }$