Question

2
In the figure , LABC = 90'
and
seg
BO I side Ac and
A.D.C.. then by theorem
of Geametric mean
B​

Answer 1

Answer:

BD² = AD * CD

Step-by-step explanation:

∠ABC = 90°

BD ⊥ AC

in Δ ADB  & ΔABC

∠ADB = ∠ABC = 90°

∠A = ∠A  ( common)

=> Δ ADB  ≈ ΔABC

=> AD/AB  = BD/BC  = AB/AC

AD/AB  = BD/BC

=> BD = AD * BC/AB    Eq1

in Δ BDC  & ΔABC

∠BDC = ∠ABC = 90°

∠C = ∠C  ( common)

=> Δ BDC  ≈ ΔABC

=> BD/AB  = BC/AC  = CD/BC

BD/AB  =  CD/BC

=> BD = AB * CD /BC   Eq 2

Eq1 * Eq2

=> BD² = (AD * BC/AB)  * (AB * CD/BC)

=> BD² = AD * CD