Question

In a isosceles triangle , prove that the altitude from the vertex bisects the base.

Answer 1

Step-by-step explanation:

Given:

• In an isosceles triangle (ABC).
• AD is altitude to base BC , meeting BC at D.
• AB = AC ( Two sides are equal )

To Prove:

• Altitude bisects the base of ∆ABC i.e BD = DC

Proof: In ∆ABD and ∆ADC ,

• AB = AC ( Given )

• ∠ABD = ∠ACD ( Angles opposite to equal sides are also equal )

• ∠ADB = ∠ADC ( 90 ° each )

∴ ∆ABD ≅ ∆ACD

Hence, BD = DC ( By CPCT )

$\large\boxed{\texttt{Proved}}$

_________________________

★ Another method ★

Proof: In ∆ABD and ∆ADC

• AB = AC ( given )

• AD = AD ( Common in both sides )

• ∠ABD = ∠ACD ( angles opposite to equal sides are also equal )

∴ ∆ABD ≅ ∆ACD by SAS criteria.

Hence, BD = DC ( By CPCT )

$\large\boxed{\texttt{Proved}}$

Answer 2

In an isosceles triangle (ABC).

AD is altitude to base BC , meeting BC at D.

AB = AC ( Two sides are equal )

To Prove:

Altitude bisects the base of ∆ABC i.e BD = DC

Proof: In ∆ABD and ∆ADC ,

AB = AC ( Given )

∠ABD = ∠ACD ( Angles opposite to equal sides are also equal )

∠ADB = ∠ADC ( 90 ° each )

∴ ∆ABD ≅ ∆ACD

Hence, BD = DC ( By CPCT )

Proved

_________________________

★ Another method ★

Proof: In ∆ABD and ∆ADC

AB = AC ( given )

AD = AD ( Common in both sides )

∠ABD = ∠ACD ( angles opposite to equal sides are also equal )

∴ ∆ABD ≅ ∆ACD by SAS criteria.

Hence, BD = DC ( By CPCT )