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If ABCD is a quadrilateral in which AB+CD=BC+AD,prove that the bisectors of the angles of the quadrilateral meet in a point which is equidistant from the sides of quadrilateral.

Plz provide me the proof.

Note by Brilliant Member
2 years, 1 month ago

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Connect the angle bisectors of angle A and B. Call the intersection T. Draw perpendiculars down to AB, AD, DC , CB at points X,Y,Z,W. By properties of angle bisectors, we know that \(AX=AY = a, BX = BW = b.\) Additionally, TY=TX=TW=g. Let DY, DZ, ZC, CW, be c, d, e, f respectively. Note that from the condition, \[c+f=d+e. \]By Pythagorean theorem, \[TZ= x^2+c^2-d^2=x^2 + f^2 - e^2.\]From these two equations you get that \(c = d\) and \(e = f\) which implies that DT and TC are angle bisectors which completes our proof.

Alan Yan - 2 years, 1 month ago

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What have you tried? Have you shown that the 4 angle bisectors meet at a unique point? That seems to be implicit in the problem.

Hint: Any point on the angle bisector is equidistant from both lines. Thus, once you've answered the uniqueness of intersection, you are done.

Calvin Lin Staff - 2 years, 1 month ago

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Thanks for help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Really Glad !

Brilliant Member - 2 years, 1 month ago

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