Theorem 6.7: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then right triangle on both sides of the perpendicular are similar to the whole triangle and to each other Given: ∆ABC right angled at B
& perpendicular from B intersecting AC at D. (i.e. BD ⊥ AC)
To Prove: ∆ADB ~ ∆ABC
∆BDC ~ ∆ABC
& ∆ADB ~ ∆BDC
Theorem 6.7: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then right triangle on both sides of the perpendicular are similar to the whole triangle and to each other Given: ∆ABC right angled at B
& perpendicular from B intersecting AC at D. (i.e. BD ⊥ AC)
To Prove: ∆ADB ~ ∆ABC
∆BDC ~ ∆ABC
& ∆ADB ~ ∆BDC
Proof:
In ∆ADB & ∆ABC
∠ A = ∠ A
∠ ADB = ∠ ABC
∆ADB ~ ∆ABC
Similarly,
In ∆BDC & ∆ABC
∠ C = ∠ C
∠ BDC = ∠ ABC
∆BDC ~ ∆ABC
From (1) and (2)
∆ADB ~ ∆ABC & ∆BDC ~ ∆ABC
If one triangle is similar to another triangle, and second triangle is similar to the third triangle,
then first and third triangle are similar
∴ ∆ADB ~ ∆BDC
Hence Proved
Rough
This is same as
a = b, b = c
then a = c

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.