Question

In fig. 8.96
If angle 1 = angle 2 ,prove that BA > BD.

Answer 1

1.In ΔADC

 ∠ADB is an exterior angle for ΔADC

∠ADB = ∠DAC + ∠ACD (from exterior angle property)

 Now we can also write

 ∠ADB > ∠DAC  ...1)

2. But it is given that

   ∠BAD =∠DAC    ...2)

3. From equation 1) and equation 2)

   ∠ADB > ∠BAD    ...3)

4. Now in ΔBAD

   where

   ∠ADB > ∠BAD

   So

   BA > BD     (because side opposite to larger angle is larger)

Answer 2

Answer:

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