how do you prove that a quadrilateral is a kite?
By definition of kite, you'd like to show that it has two pairs of adjacent sides which are congruent.
so if you have a quad BIRD and ID bisects RB and BI is congruent to IR how do u prove its a kite..?
Let me see.
ohkay thankyou.
Dude sorry for the delay my computer's so lagging like crazy. First, you can show that IRB is an isosceles triangle with vertex I. Then if ID bisects RB, it follows that ID is perpendicular to RB since if you drop the bisector down from the vertex, that bisector is perpendicular to the side it’s bisecting. So ID is perpendicular to RB. But then call intersection of ID and RB O. Then consider triangles ROD and BOD. We will show those are congruent. We know that RO = BO since RB was bisected. We also know Angle ROD = Angle BOD since both are 90 degrees. And OD = OD itself. So by SAS, the triangles are congruent. Therefore, RD = BD. So now we have RD = BD, and also IR = IB, since that triangle was isosceles. Therefore, we have a kite.
what wpuld be the reason for RO and BO to be perpendicular?! and what would be the reason for RD = BD and IR=IB... CPCTC?
RO and BO are not perpendicular. They lie on the same line segment. But ID and BR are perpendicular. This is because in an isosceles triangle, if you drop the bisector from the top vertex, then it is actually a perpendicular bisector. Yes, RD and BD are equal because CPCTC. ROD and BOD were shown to be congruent. IR and IB are congruent by assumption.
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