## Theorem of Geometric mean

**Theorem : **

**In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.**

**Given :**

In ∆ABC,

∠ABC = 90°

seg BD ⊥ seg AC

A – B – C

**To prove :**

BD² = AD × DC

**Proof :**

In ∆ADB and ∆ABC

∠DAB ≅ ∠BAC …*[Common angle]*

∠ADB ≅ ∠ABC …*[Both are 90°]*

∆ADB ~ ∆ABC …*[By AA test of similarity of triangles]* (i)

In ∆BDC and ∆ABC

∠BCD ≅ ∠ACB …*[Common angle]*

∠BDC ≅ ∠ABC …*[Both are 90°]*

∆BDC ~ ∆ABC …*[By AA test of similarity of triangles]* (ii)

∴ ∆ADB ~ ∆BDC …*[From (i) and (ii)]*

∴ \(\large \frac {BD}{DC}\) = \(\large \frac {AD}{BD}\)

∴ BD² = AD × DC

∴ BD is the geometric mean of AD and DC

**Hence Proved.**