Given:In a triangle ABC in which EF BC
To prove that:AE/EB=AF/FC
Construction:Draw EX perpendicular AC and FY perpendicular AB
Proof:taking the ratios of area of triangle AEF and EBF and second pair of ratio of area of triangle AEF and ECF.
We get AE/EB and AF/FC
we know that triangle lie b/w sme and same base is equal in area
therefore area of EBF I equal to area of ECF
therefore AE/EB=AF/FC
HENCE PROVED
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Two congruent triangles.. To prove it, use the SSS Postulate.
No. You can know all three angles of both and all you can say is that the triangles are similar. Or with any pair of congruent sides you can have an acute angle between them or an obtuse angle.
Postulates are assumed to be true and we need not prove them. They provide the starting point for the proof of a theorem. A theorem is a proposition that can be deduced from postulates. We make a series of logical arguments using these postulates to prove a theorem. For example, visualize two angles, two parallel lines and a single slanted line through the parallel lines. Angle one, on the top, above the first parallel line is an obtuse angle. Angle two below the second parallel line is acute. These two angles are called Exterior angles. They are proved and is therefore a theorem.
Two triangles are similar if: 1) 3 angles of 1 triangle are the same as 3 angles of the other or 2) 3 pairs of corresponding sides are in the same ratio or 3) An angle of 1 triangle is the same as the angle of the other triangle and the sides containing these angles are in the same ratio. So if they are both equilateral, then they both have three 60 degree angles since equilateral triangles are equiangular as well. Then number 1 above tell us by AAA, they are similar.
The first thing you prove about congruent triangles are triangles that have same side lines (SSS) is congruent. (some people DEFINE congruent that way). You just need to show AAS is equivalent or implies SSS and you are done. That's the first theorem I thought of, don't know if it works though, not a geometry major.