ABC and DBC are both isosceles triangles on a common base BC such…

Question

ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If $\angle\text{BAC}=40^\circ$ and $\angle\text{BDC}=100^\circ,$then find $\angle\text{ADB}.$

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Answer

YES $\triangle\text{ADB}\cong\triangle\text{ADC}$ (By SSS)
AB = AC , DB = DC and AD= DA
$\angle\text{BAD}=\angle\text{CAD}$ (C.P.C.T)
$\angle\text{BAD}+\angle\text{CAD}=40^\circ$
$\Rightarrow2\angle\text{BAD}=40^\circ$
$\Rightarrow\angle\text{BAD}=\frac{40^\circ}{2}$
$\angle\text{BAD}=20^\circ$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{BAC}=180^\circ$ (Angle sum property)
Since $\triangle\text{ABC}$ is an isosceles triangle,
$\angle\text{ABC}=\angle\text{BCA}$
$\angle\text{ABC}+\angle\text{ABC}+40^\circ=180^\circ$
$\Rightarrow2\angle\text{ABC}=180^\circ-40^\circ$
$\Rightarrow2\angle\text{ABC}=140^\circ$
$\Rightarrow\angle\text{ABC}=\frac{140^\circ}{2}$
$\angle\text{ABC}=70^\circ$
$\angle\text{DBC}+\angle\text{BCD}+\angle\text{BDC}=180^\circ$ (Angle sum property)
Since $\triangle\text{ABC}$ is an isosceles triangle,
$\angle\text{DBC}=\angle\text{BCD}$
$\angle\text{DBC}+\angle\text{DBC}+100^\circ=180^\circ$
$\Rightarrow2\angle\text{DBC}=180^\circ-100^\circ$
$\Rightarrow2\angle\text{DBC}=80^\circ$
$\Rightarrow\angle\text{DBC}=\frac{80^\circ}{2}$