ABCD is a convex quadrilaterial such that AC=BD. AC and BD intersect at E and ∠AEB=68∘. F and G are the midpoints of AD and BC, respectively. FG intersects AC and BD at H and I, respectively. What is the measure (in degrees) of ∠EHI?
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i) Wish you have sketch prepared for this question. Due to some constraints, it is not possible for me to upload a sketch for the same.
ii)Mark mid points P & Q on the sides AB & CD respectively. Join PGQFP in the order, which is a Rhombus. [PG & QF parallel to AC and half of AC; similarly GQ & FP are parallel to BD and half of BD; as AC = BD, all these four sides are equal; hence it is rhombus].
iii) As from the above AC parallel FQ and BD parallel QG, <FQG = <HEI = 68⁰
So from the triangle FQG, <QFG = <QGF = (180 - 68)/2 = 56⁰
Again <EHI = <QFG = 56⁰ [As AC parallel FG, corresponding angles are equal]
Thus measure of <EHI = 56⁰