Question

in figure 6.6 ,prove that angle ADC=angle A+angle B+angle C

Answer 1

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Answer 2

Solution-

Extending AC we get 2 triangles, i.e ΔABC and ΔADC.

From the properties of triangle, the sum of all the 3 angles of a triangle is 180°

So in ΔABC,

$\Rightarrow \angle ACB+\angle CBA+\angle BAC=180$

And

$\angle ACB=\angle ACD+\angle DCB\ and\ \angle BAC=\angle DAC+\angle DAB$

Again in ΔADC,

$\Rightarrow \angle ACD+\angle CDA+\angle DAC=180$

Comparing both the equations,

$\Rightarrow \angle ACB+\angle CBA+\angle BAC=\angle ACD+\angle CDA+\angle DAC$

Substituting the values,

$\Rightarrow \angle ACD+\angle DCB+\angle CBA+\angle DAC+\angle DAB=\angle ACD+\angle CDA+\angle DAC$

$\Rightarrow \angle DCB+\angle CBA+\angle DAB=\angle CDA$

$\Rightarrow \angle CDA=\angle C+\angle B+\angle A$

$\Rightarrow \angle ADC=\angle A+\angle B+\angle C\ (Proved)$