Question

ABD is a triangle right angled at A and AC is perpendicular to BD.
SHOW That 1. AB²=BC×BD
2.AC²=BC×DC
3.AD²=BD×CD​

Answer 1

Answer:

Step-by-step explanation:

Given :

In ∆ADB ,

<A= 90° and AC perpendicular to BD

Prove : i ) AB² = BC . BD

ii ) AC² = BC . DC

iii ) AD² = BD . CD

Proof :

i ) From ∆ABD , ∆ACB

<B is common angle -----( 1 )

<DAB = <ACB = 90°-------( 2 )

From ( 1 ) and ( 2 ) ,

∆ABD ~ ∆CBA ( A.A similarity )

DA/AC = DB/AB = AB/BC

BD/AB = AB/BC

=> AB² = BD . BC ------( i )

ii ) From ∆DAB , ∆DAC

<D is common angle

<DAB = <DCA = 90°

Therefore ,

∆DAB ~ ∆DCA ( A.A similarity )

DA/DC = DB/DA

AD/CD = BD/AD

=> AD² = BD .CD -----( iii )

iii ) Since ∆ABD ~ ∆ACB and

∆DAB ~ ∆DCA

=> ∆ACB ~ ∆DCA

then ,

DC/AC = DA/AB = AC/BC

=> AC² = DC . BC --------( ii )

I hope this helps you.

: )