Question

in traingle abc if angle c greater than ane b then
a)bc greater than ac
b)bc lesser than ac
c) ab greater than ac
d)ab lesser than ac​

Answer 1

Answer:

option dddddddddddddddddddddddddd

Answer 2

Step-by-step explanation:

In a triangle, the length of any side is less than the sum of the other two sides.

So in a triangle ABC, |AC| < |AB| + |BC|. (Also, |AB| < |AC| + |CB|; |BC| < |BA| + |AC|.)

This is an important theorem, for it says in effect that the shortest path between two points is the straight line segment path. This is because going from A to C by way of B is longer than going directly to C along a line segment.

Proof:

We will add something to the figure that “straightens out” the broken path. Construct point D on line AB so that BD = BC and so |AD| = |AB| + |BC|.

Then since triangle BDC is isosceles by construction of D, then the base angles DCB and CDB are congruent. But angle DCB is smaller than angle DCA; for this angle is contained inside angle DCA, since B is between D and A.

But this means that in the triangle ADC, Angle D is less than angle C, so for the opposite sides: |AC| < |AD|. But |AD| = |AB| + |BC| by the ruler axiom.

Triangle equality” and collinearity

Theorem: If A, B, C are distinct points in the plane, then |CA| = |AB| + |BC| if and only if the 3 points are collinear and B is between A and C (i.e., B is on segment AC).

Proof:

First we prove that the equality is true if B is between A and C. Choose a ruler on the line AB; then the 3 points correspond to numbers a, b, c and either a < b < c or c < b < a.

Suppose a < b < c. Then |AC| = |c – a| = c – a = (c – b) + (b – a) = |c – b| + |b – a| = |AB| + |BC|. This uses (carefully) the order of the numbers, so that |c – a| = c – a, because a < c, for example. The case c > b > a gives the same result.

For the other direction, the converse, we must prove that if |AC| = |AB| + |BC|, then the points are collinear and B is between A and C.

First, the points must be collinear, for if they were not, then ABC would be a triangle and the triangle inequality would be true. If the points are collinear, then as we saw from the ruler computation, B must be between A and C. If this were not the case, and if, say A were between B and C, then |CB| would = |CA| + |AB|. But we would still have |CA| = |AB| + |BC|, so together these would give |CB| = |AB| + |BC| + |AB|, or 0 = 2|AB|. This is impossible for distinct points A and B.