Question

In TRIANGLE ABC AD PERPENDICULAR TO BC SUCH THAT AD SQUARE IS EQUAL TO BD*CD PROVE TRIANGLE ABC RIGHT ANGLED AT A

Answer 1

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.

Answer 2

Answer:

Step-by-step explanation: