Question

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

Answer 1

Given :-

• D is any point on AC of a triangle ABC

• AB = AC

To prove :-

CD < BD

Proof :-

Here, In triangle ABC

AB = AC ( Given )

Therefore,

AngleABC = AngleACB. ( 1 )

[Angle opposite to equal sides are equal]

Now,

In ΔABC and ΔDBC

Angle ABC > Angle DBC

[Since, Angle DBC is an internal angle of Angle B]

Angle ACB > Angle DBC ( From ( 1 ) )

BD > CD

[ Side opposite greater angle is longer ]

CD < BD $.$

Answer 2

Answer:

Given :-

• D is any point on AC of a triangle ABC

• AB = AC

To prove :-

CD < BD

Proof :-

Here, In triangle ABC

AB = AC ( Given )

Therefore,

AngleABC = AngleACB. ( 1 )

[Angle opposite to equal sides are equal]

Now,

In ΔABC and ΔDBC

Angle ABC > Angle DBC

[Since, Angle DBC is an internal angle of Angle B]

Angle ACB > Angle DBC ( From ( 1 ) )

BD > CD

[ Side opposite greater angle is longer ]

CD < BD

$\:$