Thứ Bảy, 13 tháng 9, 2014

1- Let $ABCD$ be a quadrilateral, $P$ be a point on the plane. If $\angle APB+\angle CPD=180^0$ then have a $P'$ isogonal conjugates of $P$ . (and some result another)

http://www.geogebratube.org/student/m81105

2- If $\angle APB+\angle CPD=180^0$ and $PA.PC=PB.PD$ then center of Pedal circles is Midpoints of two midpoint of diagonals

3-If $\angle APB+\angle CPD=180^0$ and $PA.PC=PD.PB$ , $M$ is midpoint of $AD$ , $M$ meets $BC$ at $K$ . Then $\angle MKC = \angle DPC$ (a generalization Brahmagupta's theorem)

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Thứ Bảy, 13 tháng 9, 2014

45-Pedal circle of a quadrilateral

1- Let $ABCD$ be a quadrilateral, $P$ be a point on the plane. If $\angle APB+\angle CPD=180^0$ then have a $P'$ isogonal conjugates of $P$ . (and some result another)

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Thứ Bảy, 13 tháng 9, 2014

45-Pedal circle of a quadrilateral

1- Let be a quadrilateral, be a point on the plane. If then have a isogonal conjugates of . (and some result another)

Không có nhận xét nào:

1- Let $ABCD$ be a quadrilateral, $P$ be a point on the plane. If $\angle APB+\angle CPD=180^0$ then have a $P'$ isogonal conjugates of $P$ . (and some result another)