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  • D is any point on side AC of a Triangle ABC with AB equal to AC. Show that CD less than BD.

D is any point on side AC of a \triangleABC with AB = AC. Show that CD < BD.

Answers (1)

Given: ABC is a triangle and D is any point on AC.

AB = AC

To prove: CD < BD.

Proof: In \triangleABC

AB = AC                                             (Given)

\angleABC = \angleACB                      (angles opposite to equal sides are equal)

In \triangleABC and \angleDBC

\angleABC > \angleDBC

\Rightarrow \angleACB > \angleDBC

\Rightarrow BD > CD                            (Side opposite to greater angle is larger)

\therefore CD < BD

Hence proved.

Posted by

infoexpert24

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