In the given figure,∆ABc is a right angled at B such that ∠BCA = 2

1.	In the given figure,∆ABc is a right angled at B such that ∠BCA = 2∠BAC. Show that hypotenuse AC = 2BC
Answer» $\bf\underline{\underline{\pink{Question:-}}}$ ★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

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

Answer»

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

★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