Question

ABD is a triangle right angled at A and AC prependicular to BD. Show that: AB^2= BC×BD ​

Answer 1

In ΔADB and ΔCAB, we have

∠DAB = ∠ACB (Each equals to 90°)

∠ABD = ∠CBA (Common angle)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒ AB/CB = BD/AB

⇒ AB2 = CB × BD

Answer 2

Hey mate!!

__________________________________________________________

Given,

ABD is right angled ∆

Where,

AC is perpendicular to BD

Now,

→ (i) In ΔADB and ΔCAB,

we have

∠DAB = ∠ACB (Each equals to 90°)

∠ABD = ∠CBA (Common angle)

∴ ΔADB ~ ΔCAB [AA similarity

criterion]

→ AB/CB = BD/AB

→ AB2 = CB × BD

Hope it helps!!

# Jaihind