Question

Hi Friends,

Prove that ∆ ABC is isosceles if altitudes AD bisect BC.

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Class - 9th
Chapter - Triangle

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Answer 1

Altitude AD bisects ∠BAC

In △BAD and △CAD

AD=AD (Common Side)

∠ADB=∠ADC=90

o

(AD is altitude)

∠BAD=∠CAD (AD bisects ∠BAC)

∴△BAD≅△CAD(A.S.A)

⇒AB=AC(C.P.C.T.C.)

Thus, △BAC is an isosceles triangle.

B)

Median AD is perpendicular to BC

In △BAD and △CAD

AD=AD (Common Side)

∠ADB=∠ADC=90

o

(AD is perpendicular to BC)

BD=CD (AD is Median)

∴△BAD≅△CAD(S.A.S.)

⇒AB=AC(C.P.C.T.C.)

here's your answer

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Answer 2

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Solution :

Given :- ∆ABC , in which altitude AD bisect BC.

To prove :- ∆ABC is Isosceles.

Proof :- In ∆ ABD and ∆ ACD ,

We have ,

⇒BD = DC(Given)

⇒∠ ADB = ∠ADC ( Each = 90°)

⇒AD = AD ( Common side)

⇒∆ ABD ≌ ∆ ACD ( By SAS )

⇒AB = AC ( ∵ C.P.C.T )

∴ ∆ ABC is an isosceles triangle Proved.

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