Question

In the figure $AB=AC,BD=DC,\angle C={40}^o\angle BDC={160}^o$
Write the measure of all angles of triangle $ABD$.

Answer 1

Consider the following question.

Given that.

$AB=AC,BD=DC,\angle C={40}^0\angle BDC={160}^0$

Find the value of $\Delta ABD=?$

We know that,

$\angle BAC=\frac{1}{2}\angle BDC$

$=\frac{1}{2}\times {160}^0$

$\angle BAC={80}^0$

In $\Delta CAD$,

Therefore,

$\angle ADC+\angle DCA+\angle CAD={180}^0$

$\angle ADC+{40}^0+{40}^0={180}^0$

$\angle ADC={100}^0$

In $\Delta ABD$

$\angle BAD=360-(\angle BDC+\angle CDA)$

$={360}^0-({160}^0+{100}^0)$

$\angle BAD={100}^0$

$\angle DAB={40}^0$

$\angle BAD+\angle DAB+\angle ABD={180}^0$

${40}^0+{100}^0+\angle ABD={180}^0$

$\angle ABD={40}^0$

Hence, this is the required answer.