Question

Question No.1

ABC is a right angled triangle. AD is perpendicular to the hypotenuse BC. If AC = 2AB, prove that BD = BC / 5

With Steps Please

Answer 1

We are given with AC = 2AB

and AD is perpendicular to BC where BC is the hypotenuse.

now in Δ ABC (AB²) + (AC)² = (BC)²

As, AC = 2 AB therefore, (2AB)² + (AB)² = (BC)²

= 5 (AB)² = (BC)² --- (1)

now we can write the area of triangle  either as 1/2 (AB) (AC) or as 1/2 (BC) (AD)

When we equate both 
(AB) (AC) = (BC) (AD)
 as AC = 2 (AB)
= 2 (AB)² = (BC)(AD)
= (AD) = 2 (AB)² / (BC)
= (AD)² = 4 (AB)^4 / (BC)²

Now from (1)

(BC)² = 5 (AB)²
= (AD)² = 4/5 (AB)² --- (2)

Now in Δ ABD,

(BD)² = (AB)² - (AD)²

From (2),

(BD)² = (AB)² - 4/5 (AB)² = 1/5 (AB)² --- (3)

and from (1) (BC)² = 5 (AB)² --- (4)

Now on the last step 

We will take square.

from (3) and (4)

(BD/BC)² = 1/5 (AB)² / 5 (AB)² = 1/25

so when we apply the square root to both sides we will get

(BD)/ (BC)  = 1/5

= BD = 1/5 BC