∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that
(i) AB² = BC . BD
(ii) AC² = BC . DC
(iii) AD² = BD . CD
(iv) AB²/AC² = BD/DC
Answers
SOLUTION :
We know that if a perpendicular is drawn from the vertex of the right angle of a right angled triangle to the hypotenuse, then triangle on both sides of the perpendicular are similar to the whole triangle and to each other.
So , ∆ACB ~ ∆DAB ~ ∆DCA ……….(1)
(i) From eq 1,
∆ACB ~ ∆DAB
Therefore, ar(∆ACB)/ar(∆DAB) = AB²/DB²
[The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides]
½ × BC × AC/½ × BD × AC = AB²/DB²
[area of triangle = ½ × base × height]
BC/BD = AB²/DB²
AB² = BC × BD …………..(2)
(ii) From eq 1,
∆ACB ~ ∆DCA
Therefore, ar(∆ACB)/ar(∆DCA) = AC²/DC²
[The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides]
½ × BC × AC/½ × DC × AC = AC²/DC²
[area of triangle = ½ × base × height]
BC/DC = AC²/DC²
AC² = BC × DC ……………(3)
(iii) From eq 1,
In ∆DCA ~ ∆DAB
Therefore, ar(∆DCA) /ar(∆DAB) = DA²/DB²
[The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides]
½ × CD × AC/½ × BD × AC = AD²/DB²
[area of triangle = ½ × base × height]
CD/BD = AD²/BD²
AD² = BD × CD ………(4)
(iv) On dividing eq 2 & 3
AB² /AC² = BC /BC × BD/DC
AB² /AC² = BD/DC
HOPE THIS ANSWER WILL HELP YOU..
Answer:
ac^2=ad^2-dc^2 in triangle adc
ac^2=ab^2-bc^2 in triangle acb
add the above
2ac^2=ad^2+ab^2-dc^2-bc^2
=bd^2-dc^2-bc^2
=(bc+cd)^2-dc^2-bc^2
=bc^2+cd^2+2bc*cd-dc^2-bc^2
=2bc*cd
so ac^2 =bc*cd
this is based on pythogorus theorem.