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.
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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=x2+c2−d2=x2+f2−e2.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.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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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=x2+c2−d2=x2+f2−e2.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.
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.
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