Question

Angle A = 90 , AD perpendicular to BC . If BD = 2 cm and CD = 8 cm, find AD.

Answer 1

In ∆ ABC, we have, ∠A = 90°.
Also, AD⊥BC.
Now,In ∆ABC, we have,
∠BAC = 90°⇒∠BAD + ∠DAC = 90° .....(1)
In ∆ADC, we have,
∠ADC = 90°
So, ∠DCA + ∠DAC = 90° ......(2)
from (1) and (2), we have
∠BAD + ∠DAC = ∠DCA + ∠DAC
⇒∠BAD = ∠DCA ......(3)
In ∆BDA and ∆ADC,
∠BDA = ∠ADC (90° each)
∠BAD = ∠DCA (proved above)
so, ∆BDA ~ ∆ADC (AA similarity)

⇒BD/AD = AD/DC = AB/AC (Corresponding sides of similar ∆'s are proportional)
⇒BD/AD = AD/DC
⇒AD^2 = BD × CD
⇒AD^2 = 2 × 8 = 16
⇒AD = 4 cm

Answer 2

Answer:

In ∆ ABC, we have, ∠A = 90°.

Also, AD⊥BC.

in ∆ABC, we have,

∠BAC = 90°⇒∠BAD + ∠DAC = 90° -----1

In ∆ADC, we have,

∠ADC = 90°

So, ∠DCA + ∠DAC = 90° -------2

from (1) and (2), we have

∠BAD + ∠DAC = ∠DCA + ∠DAC

⇒∠BAD = ∠DCA -----3

In ∆BDA and ∆ADC,

∠BDA = ∠ADC (90° each)

∠BAD = ∠DCA (proved above)

so, ∆BDA ~ ∆ADC (AA similarity)

⇒BD/AD = AD/DC = AB/AC (Corresponding sides of similar ∆'s are proportional)

⇒BD/AD = AD/DC

⇒AD^2 = BD × CD

⇒AD^2 = 2 × 8 = 16

⇒AD = 4 cm