Question

In the figure angle abc equal to 90 degree and seg BD perpendicular side AC and mean BD square is equal to in the figure angle abc 90 degree and BD perpendicular side AC and ADC then by theorem of geometric mean BD square

Answer 1

BD^2=AD×DC is the answer of the question

Answer 2

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

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