In quadrilateral ABCD, B is reflection of D in AC and AE bisects angle CAD such that C is on CD,Angle DEA is 3 times angle DAE, PROVE BCD is

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In quadrilateral ABCD, B is reflection of D in AC and AE bisects angle CAD such that C is on CD,Angle DEA is 3 times angle DAE, PROVE BCD is a rhombus

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Ivy 1 month 2021-08-14T01:04:06+00:00 1 Answer 0 views 0

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    2021-08-14T01:05:16+00:00

    Answer:

    please mark me as breainlist answer

    Step-by-step explanation:

    AC and BD are diagonals . So let they intersect at O.

    In triangles ABC and ADC,

    angle CAB = angle CAD

    angle ACB = angle ACD

    AC = AC

    Therefore, triangles ABC and ADC are congruent.

    NOW,

    AB = AD (corresponding parts of congruent triangles)

    In triangle ABD,

    AB = AD

    So, triangle ABD is isosceles.

    As we know the angle bisector of the vertical angle of an isosceles triangle

    is also the perpendicular bisector of the base.

    So AC intersects BD at right angle.

    so angle AOD = 90o.

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