in triangle abc ad perpendicular to BC such that a b square is equal to BD into CD prove that triangle ABC is right angled at A
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Given :
In triangle ABC , AD is perpendicular to BC and AD² = BD.DC
To prove :
BAC = 90°
Proof :
When perpendicular is drawn from A, it makes 2 right angles on BC line.
(Both 90°)
Thereby, forming 2 right triangles ∆ADB and ∆ADC.
Apply Pythagoras theorem.
Hyp²= Opp² + Adj²
{In ∆ADB,
Hyp= AB ; Opp= BD ; Adj= AD
In ∆ADC,
Hyp = AC ; Opp= DC ; Adj= Ad}
Then we have ,
AB² = AD² + BD² ----------(1)
AC²= AD²+ DC² ---------(2)
Add (1)&(2)
AB² + AC² = 2(AD²) + BD²+ DC²
= 2(BD . CD) + BD² + CD²
[ ∵ given AD² = BD.CD ]
= (BD + CD )² = BC²
Thus in triangle ABC we have ,
AB² + AC²= BC²
{Here in ∆ABC, Hyp= BC, Adj= AC , Opp= AB}
so Hyp² = Adj² + Opp²
•°• Pythagoras theorem is satisfied
So triangle ABC is a right triangle right angled at A
∠ BAC = 90°
:)
Hope it helps you friend
In triangle ABC , AD is perpendicular to BC and AD² = BD.DC
To prove :
BAC = 90°
Proof :
When perpendicular is drawn from A, it makes 2 right angles on BC line.
(Both 90°)
Thereby, forming 2 right triangles ∆ADB and ∆ADC.
Apply Pythagoras theorem.
Hyp²= Opp² + Adj²
{In ∆ADB,
Hyp= AB ; Opp= BD ; Adj= AD
In ∆ADC,
Hyp = AC ; Opp= DC ; Adj= Ad}
Then we have ,
AB² = AD² + BD² ----------(1)
AC²= AD²+ DC² ---------(2)
Add (1)&(2)
AB² + AC² = 2(AD²) + BD²+ DC²
= 2(BD . CD) + BD² + CD²
[ ∵ given AD² = BD.CD ]
= (BD + CD )² = BC²
Thus in triangle ABC we have ,
AB² + AC²= BC²
{Here in ∆ABC, Hyp= BC, Adj= AC , Opp= AB}
so Hyp² = Adj² + Opp²
•°• Pythagoras theorem is satisfied
So triangle ABC is a right triangle right angled at A
∠ BAC = 90°
:)
Hope it helps you friend
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