If A,B,C,D are position vectors of the vertices of a cyclic quadrilateral ABCD then we have to prove that
I took the origin as centre of the circle .
Then magnitude of each vector is r .
Now how can I proceed .
position vectors of the vertices of a cyclic quadrilateral
3
$\begingroup$
vector-spaces
-
0And please don't keep tagging vector spaces. Use the vectors tag. – 2017-01-21
-
0Sorry , I will not do it again @Rohan – 2017-01-21
1 Answers
2
We have $$\tan A =\frac {\sin A}{\cos A} =\frac {\overrightarrow{AB} \times \overrightarrow{AD}}{\overrightarrow {AB}\cdot \overrightarrow {AD}}$$ $$ = \frac {(\vec b -\vec a)\times (\vec d - \vec a)}{(\vec b- \vec a)\cdot (\vec d-\vec a)} $$ $$= \frac {|\vec a\times \vec b+\vec b\times \vec d+\vec d \times \vec a|}{(\vec b-\vec a)\cdot (\vec d-\vec a)} $$ Similarly, we can show that the second term equals $\tan C $. We thus need to prove that $$\tan A +\tan C =0 \implies \sin (A+C)=0 \implies A+C =\pi $$ which is true as $ABCD $ is cyclic. Hope it helps.