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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.

#Geometry

Note by A Former Brilliant Member
5 years, 7 months ago

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Comments

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.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.c+f=d+e. By Pythagorean theorem, TZ=x2+c2d2=x2+f2e2.TZ= x^2+c^2-d^2=x^2 + f^2 - e^2.From these two equations you get that c=dc = d and e=fe = f which implies that DT and TC are angle bisectors which completes our proof.

Alan Yan - 5 years, 7 months ago

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 - 5 years, 7 months ago

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

A Former Brilliant Member - 5 years, 7 months ago
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