Solution:
c) BD.CD=AD²
Explanation:
From triangles ADB and ADC,
Now according to the question, we have,
∠ADB = ∠ADC = 90° (Since AD ⊥ BC)
∠DBA = ∠DAC [As each angle = 90°- ∠C]
Using AAA criterion for similarity,
∆ADC∼ ∆ADB
AD/CD = BD/AD
AD2 = BD.CD
In figure, if ∠BAC =90° and AD⊥BC. Then, (a) BD.CD = BC² (b) AB.AC = BC² (c) BD.CD=AD² (d) AB.AC =AD²
In figure, if ∠BAC =90° and AD⊥BC. Then, (a) BD.CD = BC² (b) AB.AC = BC² (c) BD.CD=AD² (d) AB.AC =AD²