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प्रश्न
In the following Figure ∠ACB= 90° and CD ⊥ AB, prove that CD2 = BD × AD
उत्तर
Given that: CD ⊥ AB
∠ACB = 90°
To prove : CD2 = BD × AD
Using pythagoras Theorem in Δ ACD
AC2 = AD2 +CD2 ..........(1)
Using pythagoras Theorem in ΔCDB
CB2 = BD2+CD2 .....(2)
Similarly in ΔABC,
As AB = AD + DB
Since ,AB = AD +BD ......(4)
Squaring both sides of equation (4), we get
(AB)2 = (AD+BD)2
Since, AB2 = AD2 +BD2 +2×BD×AD
From equation (3) we get
AC2 +BC2=AD2+BD2+2 × BD × AD
Substituting the value of AC2 from equation (1) and the value of BC2 from equation (2), we get
AD2 +CD2 +BD2 + CD2 +AD2 +BD2 +2×BD×AD
Since ,2 CD2 = 2 × BD × AD
Hence , CD2 = BD × AD
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