Let $ABCD$ is a convex quadrilateral.The internal angle bisectors of $\angle {BAC}$ and $\angle {BDC}$ meets at $P$.$\angle {APB}=\angle {CPD}$.Prove that $AB+BD=AC+CD$.
Problem
Source: BdMO National Higher Secondary 2019/9
Tags: geometry
Holmes777
04.03.2019 10:27
what is a convex quadrilateral?
chem123
04.03.2019 10:28
A quadrilateral in which all the angles are less than $180^o$
chem123
04.03.2019 10:29
Or, a quadrilateral in which a line joining two interior points can be drawn entirely inside it.
chem123
04.03.2019 10:36
chem123 wrote: Or, a quadrilateral in which a line joining two interior points can be drawn entirely inside it. Getting this??
Olympus_mountaineer
04.03.2019 12:50
I used a really wonderful diagram to encounter this problem.If you draw the quadrilateral and all the triangles according the question,the quadrilateral must be an isosceles trapezium and all of the triangles and the trapezium will be inscribed in a single circle!!!But proving it rigourusly is not easy.
drmzjoseph
06.03.2019 04:23
Olympus_mountaineer wrote: I used a really wonderful diagram to encounter this problem.If you draw the quadrilateral and all the triangles according the question,the quadrilateral must be an isosceles trapezium and all of the triangles and the trapezium will be inscribed in a single circle!!!But proving it rigourusly is not easy. ups it's not true because you can start tof define it with an arbitrary triangle PBC and two isogonal lines PA and PD cuting the Apollonius circles The proof is easier than this Just let M and N the reflections of B and C with respect to AP and CP respectively and we get $\triangle PMC \cong \triangle PAN$ then $MC=NA=AC-AB=BD-DC$