Read e-book online Mathematics in Ancient and Medieval India PDF

By A.K.Bag

Background of arithmetic in historical and medieval India

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Example text

There are many proofs available to justify this theorem, yet we shall use just one of these methods to prove this wonderful theorem. It is perhaps easier to follow the proof looking at the left-side diagram and then verifying the validity of each of the statements in the right-side diagram. In any case, the statements made in the proof hold for both diagrams. To prove the “only if” part of the theorem, we consider figure 2-12, where we have on the left triangle ABC with a line (SR) containing A and parallel to BC, intersecting CP extended at S and BP extended at R.

In figure 2-5, for angle bisector AT a, we get 52 . For angle bisector BT b, we get . For angle bisector CT c, we get . Now multiplying these three equations, we get which (by Ceva's theorem) allows us to conclude that the angle bisectors are concurrent. CONCURRENCY OF THE MEDIANS OF A TRIANGLE The proof that the medians are concurrent using Ceva's theorem is the simplest of all. We can see from the fact that each side is divided into two equal segments that the products of the alternate segments will be equal.

To justify it requires two proofs—the original statement and its 47 converse. For us to accept this theorem it would be appropriate to prove its validity. Ceva's theorem states the following: The three lines containing the vertices of triangle ABC (figure 2-12) and intersecting the opposite sides in points L, M, and N, respectively, are concurrent if and only if . There are many proofs available to justify this theorem, yet we shall use just one of these methods to prove this wonderful theorem.