Question

In ΔABC, ∠ACB = 90°. seg CD ⊥ side AB and seg CE is angle bisector of ∠ACB.

Prove that: `(AD)/(BD) = (AE^2)/(BE^2)`.

Answer 1

Given, In ΔACB,

m∠ACB = 90°

∵ seg CD ⊥ hypo AB

∴ seg CE is angle bisector ∠ACB.

To prove that: `(AD)/(BD) = (AE^2)/(BE^2)`

Proof: In ΔABC

seg CE is angle bisector of ∠ACB.  ......(Given)

∴ `(AC)/(BC) = (AE)/(BE)`  ......[Angle bisector theorem]

Squaring on both the sides

`(AC^2)/(BC^2) = (AE^2)/(BE^2)`  ......(i)

In ΔACB , m∠ACB = 90°  

And seg CD ⊥ hypotenuse AB ......(Given)

∴ CD² = AD × BD  ......[Geometric mean thorem]

Dividing both the sides by BD²

`(CD^2)/(BD^2) = (AD xx BD)/(BD^2)`

`(CD^2)/(BD^2) = (AD)/(BD)`  .......(ii)

Now, in the figure

ΔACB ∼ ΔADC ∼ ΔCDB  ......(Right-angled triangle similarity property)

Consider,

ΔADC ∼ ΔCDB

`(AC)/(BC) = (DC)/(BD)`  ......(c.s.c.t)

`(AC^2)/(BC^2) = (DC^2)/(BD^2)`

∴ `(AC^2)/(BC^2) = (AE^2)/(BE^2) = (AD)/(BD)` .......[From equation (i), (ii) and (iii)]

∴ `(AD)/(BD) = (AE^2)/(BE^2)`