Question

in the given figure angleA=90degree AD perpendicular to BC BD=2 cm and CD=8 cm find AD

Answer 1

ΔABC is similar to Δ DBA

Proof:

∠BAC = ∠BDA (both are right angle, given)

∠ABC = ∠DBA (common angle)

By property AA,  ΔABC is similar to ΔDBA

Find AB:

Since ΔABC is similar to ΔDBA

BC/AB = AB/BD

10/AB = AB/2

AB² = 20

AB = √20

Find AD:

ΔDBA is a right angle triangle

a² + b² = c²

AD² + 2² = (√20)²

AD² + 4 = 20

AD² = 16

AD = √16

AD = 4 cm

Answer: AD = 4 cm

Answer 2

In triangle ABC, angle A = 90 degree

AD perpendicular to BC

In triangle ABC, angle BAC = 90 degree

Angle BAC + Angle DAC = 90 eq(1)

Triangle ADC

Angle ADC = 90 degree

So, Angle DCA + Angle DAC = 90 degree eq(2)

From eq(1) and eq(2)

Angle BAD + Angle DAC = Angle DCA + Angle DAC

Angle BAD = Angle DCA eq(3)

In triangle BDA and triangle ADC,

Angle BDA = Angle ADC = 90 degree

Angle BAD = Angle DCA

So the triangle are congruent with AA

CPCT, BD/AD = AD/DC = AB/AC

BD/AD = AD/DC

AD^2 = BD * CD

AD^2 = 2 * 8 = 16

AD = 4 cm