prove that prove that sum of any two sides is greater than third side
Answers
Answer:
.ANSWER
Construction: In \Delta ABCΔABC, extend ABAB to D in such a way that AD = ACAD=AC.
In \Delta DBCΔDBC, as the angles opposite to equal sides are always equal, so,
\angle ADC = \angle ACD∠ADC=∠ACD
Therefore,
\angle BCD > \angle BDC∠BCD>∠BDC
As the sides opposite to the greater angle is longer, so,
BD > BCBD>BC
AB + AD > BCAB+AD>BC
Since AD = ACAD=AC, then,
AB + AC > BCAB+AC>BC
Hence, sum of two sides of a triangle is always greater than the third side.
hope its help.....
plzzzzzz mark it as brainliest......
Answer:
Construction: In ABCΔABC, extend ABAB to D in such a way that AD = ACAD=AC.
In DBCΔDBC, as the angles opposite to equal sides are always equal, so,
\angle ADC = \angle ACD∠ADC=∠ACD
Therefore,
\angle BCD > \angle BDC∠BCD>∠BDC
As the sides opposite to the greater angle is longer, so,
BD > BCBD>BC
AB + AD > BCAB+AD>BC
Since AD = ACAD=AC, then,
AB + AC > BCAB+AC>BC
Hence, sum of two sides of a triangle is always greater than the third side.