Question

in a right angled triangleABC, AD is perpendicular to the hypotenuse BC, then AD^2 is...............
A) AC*CD
B)AC*BA
C)DC*BD
D)None Of These

Answer 1

Answer:

Let the right angled triangle be ∆ABC right angled at A.

Let the perpendicular drop down to hypotenuse BC from point A to point D on the hypotenuse .

and, let the height of that perpendicular be 'd'.

and, let the distance of point D and between A be x and distance of point D and between B be y.

Now,

We know that this perpendicular will divide the entire triangle into two triangles which will be similar in some order.

By, considering any one angle as theta for example Angle C, you can then prove easily by simple trigonometry in the two triangles that

d^2 = x*y

(Hint : Also, angle in the other triangle i.e angle B will be 90-theta)

And ,hence the answer to this question will be that

Ad^2 = Bd*cD

Hope this helps you ! ｡◕‿◕｡

Answer 2

Answer:

here ABC is right angled triangle with <A=90 and <ADB=90

In ABDand ADC

<ADB= <ADC=90

AD= AD (common)

let <DAC= a

So, <DCA = 90-a

Also, <BAC=90

<BAD+ DAC =90

<BAD =90 -a

hence <DCA = <BAD=90-a

So, ABD ~CAD ( ASA)

Hence sides are proportional

BD/AD =AD/CD

AD^2 = BD. CD

SO, option c) is correct

#answerwithquality #BAL