Question

Prove that triangle ABC is isosceles if any one of following holds.

(1) altitude AD bisects .
(2) median AD is perpendicular to the base BC.




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Answer 1

$\huge{\textsf{\textbf{\color{pink}{Question:-}}}}$

Prove that ∆ABC is isosceles if any one of the following holds :

1. Altitude AD bisects <BAC
2. Median AD is perpendicular to the base BC.

$\huge{\textsf{\textbf{\color{lightblue}{Solution :-}}}}$

1. Altitude AD bisects <BAC :

In ∆BAD and ∆CAD

AD = AD (Common side)

<ADB = <ADC = 90° (AD is the altitude)

<BAD = <CAD (AD bisects <BAC)

∴ ∆BAD ≅ ∆CAD (By ASA rule)

==> AB = AC (CPCT)

Thus, ∆ABC is an isosceles triangle.

2. Median AD is perpendicular to the base BC :

In ∆BAD and ∆CAD

AD = AD (Common side)

<ADB = <ADC = 90° (AD is perpendicular to BC)

BD = CD (AD is median)

∴ ∆BAD ≅ ∆CAD (By SAS rule)

==> AB = AC (CPCT)

Thus, ∆ABC is an isosceles triangle.

➡️SitaVeronica✨

Hope it is helpful to uh

@AkashMello !!

Answer 2

Given :-

(1) In ∆ ABC altitude AD bisects BC.

(2) AD is perpendicular to BC.

To prove :-

∆ ABC is an isosceles ∆.

Proof :-

In ∆ ABD and ∆ ACD :-

AD = AD (common)

BD = CD (AD bisects BC)

<ADB = <ADC (each are of 90° because AD is perpendicular to BC)

so,

∆ ABD ≈ ∆ ACD (SAS)

AB = AC (by CPCT)

now,

In ∆ ABC

AB = AC

so, ∆ ABC is an isosceles ∆.

I hope it helps uh ra Mello ✨