Question

In the given figure,∆ABc is a right angled at B such that ∠BCA = 2∠BAC. Show that hypotenuse AC = 2BC​

Answer 1

$\bf\underline{\underline{\pink{Question:-}}}$

★In the given figure,∆ABc is a right angled at B such that ∠BCA = 2∠BAC. Show that hypotenuse AC = 2BC

$\bf\underline{\underline{\blue{Given:-}}}$

★A ∆ABC in which ∠B = 90° and ∠BCA = 2∠BAC

$\bf\underline{\underline{\red{To Prove:-}}}$

★AC = 2BC

$\bf\underline{\underline{\green{Construction:-}}}$

★Produce CB to D such that BD = BC. Join AD.

$\bf\underline{\underline{\orange{Proof:-}}}$

Let ∠BAC = x°. Then, ∠BCA = 2x°

In ∆ABC and ∆ABD, we have

BC = BD

AB = AB

∠ABC = ∠ABD

∴ ∆ABC ≅ ∆ABD

∴ ∠CAB = ∠DAB

and AC = AD

In ∆CAD, we have

∠CAD = ∠CAB + ∠DAB = (x° + x°) = 2x°

∠ACD = ∠ACB = 2x°

But, we know that the sides opposite to equal angles are equal.

∴ ∠ACD = ∠CAD => AD = CD

From (ii) and (iii), we get

AC = CD => AC = 2BC [∵ CD = BC + BD = 2BC]

Hence, AC = 2BC

Answer 2

Question:−

★In the given figure,∆ABc is a right angled at B such that ∠BCA = 2∠BAC. Show that hypotenuse AC = 2BC

Given:-

★A ∆ABC in which ∠B = 90° and ∠BCA = 2∠BAC

ToProve:−

★AC = 2BC

Construction:−

★Produce CB to D such that BD = BC. Join AD.

Proof:-

Let ∠BAC = x°. Then, ∠BCA = 2x°

In ∆ABC and ∆ABD, we have

BC = BD

AB = AB

∠ABC = ∠ABD

∴ ∆ABC ≅ ∆ABD

∴ ∠CAB = ∠DAB

and AC = AD

In ∆CAD, we have,

∠CAD = ∠CAB + ∠DAB = (x° + x°) = 2x°

∠ACD = ∠ACB = 2x°

But, we know that the sides opposite to equal angles are equal.

∴ ∠ACD = ∠CAD => AD = CD

From (ii) and (iii), we get

AC = CD => AC = 2BC [∵ CD = BC + BD = 2BC]

Hence, AC = 2BC.