SOLUTION: QUESTION: Prove that the area of a kite is equal to half the product of its diagonals. I started by stating that ABCD is a kite with AB=AD and CD=CB and E is the intersection of

Algebra ->  Geometry-proofs -> SOLUTION: QUESTION: Prove that the area of a kite is equal to half the product of its diagonals. I started by stating that ABCD is a kite with AB=AD and CD=CB and E is the intersection of      Log On


   

Question 167941: QUESTION: Prove that the area of a kite is equal to half the product of its diagonals.
I started by stating that ABCD is a kite with AB=AD and CD=CB and E is the intersection of the lines containing the diagonals AC and BD. I showed that AC=AC by the reflexive property. The triangle ABC is congruent to triangle ADC by SSS congruence postulate. THe angle BAE is congruent to angle DAE by corresponding parts of congruent triangles are congruent. The triangle ADE is congruent to triangle ABE by SAS. The line DE is congruent to line BE by corresponding parts of congruent triangles are congruent. The Angle AED is congruent to angle AEB by corresponding parts of congruent triangles are congruent.
I don't know what to do next/ if I am even going in the right direction. Thanks!
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!


(Area of Kite) = (Area of triangle ABD) + (Area of triangle BCD)

(Area of Kite) =     1%2F2BD%2AAE%22%2B%221%2F2BD%2AEC

Factor out 1%2F2BD

(Area of Kite) = 1%2F2BD%2A%28AE%2BEC%29

%28AE%2BEC%29=AC

(Area of Kite) = 1%2F2BD%2AAC

Edwin