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

Answer 1

(1)Given : A triangle ABC  in which BD = DC

to prove : AB = AC or ABC is an isosceles triangle .
 
 Proof:
 
In triangle ABD and triangle ADC

AD = AD ( common)
∠ADB = ∠ADC (90° each)
BD = DC (given)

so Δ ABD ≡ ΔACD( by RHS)

So AB = AC (by cpct)
∴ Δ ABC is an isosceles triangle
(2)

Answer 2

$<b >Hey here is your answer$

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 ∆.

hope this helps you

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