Prove that \[\angle COD=\frac{1}{2}[\angle A+\angle B].\] |
Answer:
In\[\Delta COD\], we have \[\angle COD+\angle \text{1}+\angle 2=180{}^\circ \] \[\Rightarrow \] \[\angle COD\,=180{}^\circ -[\angle 1+\angle 2]\] \[\Rightarrow \]\[\angle COD\,=180{}^\circ -\left[ \frac{1}{2}\angle D+\frac{1}{2}\angle C \right]\] \[\Rightarrow \]\[\angle COD\,=180{}^\circ -\frac{1}{2}[\angle D+\angle C]\] But \[\angle A+\angle B+\angle C+\angle D=360{}^\circ \] \[\Rightarrow \] \[\angle C+\angle D=360{}^\circ -(\angle A+\angle B)\] \[\Rightarrow \] \[\angle COD=180{}^\circ -\frac{1}{2}[360{}^\circ -(\angle A+\angle B)]\] \[=\text{ }180{}^\circ -\frac{1}{2}\text{ }\left[ 360{}^\circ \right]\text{ }+\text{ }\frac{1}{2}\text{ }\left[ \angle A\text{ }+\text{ }\angle B \right]\] \[=\text{ }180{}^\circ -180{}^\circ \text{ }+\text{ }\frac{1}{2}\text{ }\left[ \angle A\text{ }+\text{ }\angle B \right]\] \[=\frac{1}{2}(\angle A+\angle B)\] Thus, \[\angle COD=\frac{1}{2}[\angle A+\angle B]\]
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