In Figure ABD is a triangle right angled at A and AC ⊥ BD. Show that AC^{2} = BC × DC

Advertisement Remove all ads

#### Solution

Let ∠CAB = x

In ΔCBA,

∠CBA = 180° - 90° - x

∠CBA = 90° - x

Similarly, in ΔCAD

∠CAD = 90° - ∠CBA

= 90° - x

∠CDA = 180° - 90° - (90° - x)

∠CDA = x

In ΔCBA and ΔCAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each equals to 90°)

∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]

`⇒ (AC)/(DC) = (BC)/(AC)`

⇒ AC^{2} = DC × BC

Concept: Right-angled Triangles and Pythagoras Property

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads