Question

In A triangle ABC angle ACB = 90°, BC is to produced to D such that angleDAB = 90°.

Prove that AB square = BC×BD.​

Answer 1

Step-by-step explanation:

comparing ΔABC and ΔABD

∠ACB = ∠DAB = 90° .............1st angle

∠ABC = ∠ABD = ∠B ..................2nd angle

∠BAC = ∠ADB  ........third angle will be equal too as the sum of three angle                                                                      is same (180°)

So,  ΔABC and ΔABD are similar and their similar sides will be proportional also.

$\frac{BC}{AB}=\frac{AB}{BD}=\frac{AC}{AD} = k = proportional constant$

Now, AB×AB = BC×BD

OR, AB² = BC×BD