Question

In ∆ABC, ∠ABC=90°. Seg BD ⊥ hypotenuse AC such that A-D-C

Answer 1

Answer:

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD x DC

AB² + AC² = 2AD² + BD²+ DC²

= 2BD x CD + BD² + CD² [ ∵ given AD² = BD x CD ]

= (BD + CD )² = BC²

Thus in triangle ABC we have , AB² + AC²= BC²

hence triangle ABC is a right triangle right angled at A

∠ BAC = 90°

Second Method:-

AD^2 = BD × CD

AD ÷ CD = BD ÷ AD

ΔADC ∼ ΔBDA (by SAS ∠D = 90°)

∠BAD = ∠ACD ;

∠DAC = ∠DBA (Corresponding angles of similar triangles)

∠BAD+ ∠ACD + ∠DAC + ∠DBA = 180°

⇒ 2∠BAD + 2∠DAC = 180°

⇒ ∠BAD + ∠DAC = 90°

∴ ∠A = 90°

Answer 2

Answer:

In △BCA and △BAD,

∠BCA=∠BAD        ....Each 90o

∠B is common between the two triangles.

So, △BCA∼△BAD        ...AA test of similarity     ....(I)

Hence, ABBC=ADAC=BDAB          ...C.S.S.T

And, ∠BAC=∠BDA           ....C.A.S.T        ....(II)

So, ABBC=BDAB

∴AB2=BC×BD

Hence proved.

(ii) In △BCA and △DCA,

∠BCA=∠DCA        ....Each 90o

∠BAC=∠CDA        ...From (II)

So, △BCA∼△ACD        ...AA test of similarity         ....(III)

Hence, ACBC=CDAC=ADAB          ...C.S.S.T

So, ACBC=CDAC

∴AC2=BC×DC

Hence proved.

(iii) From (I) and (III), we get

△BAD∼△ACD

Hence, ACAB<