Question

Hi xD

Triangle ABC is an isosceles triangle with AB = AC. Side BA is produced to D such that AB = AD. Prove that angle BCD is a right angle.

Acche se samjhao xD ​

Answer 1

$\huge \underline{ \rm \red{hello}}.$

Question:-

Triangle ABC is an isosceles triangle with AB = AC. Side BA is produced to D such that AB = AD. Prove that angle BCD is a right angle.

explanations:-

Given :

AB = AC

AD = AB

therefore ,

AC= AB = AD

To Prove :-

∠BCD is a Right Angle . (90°)

Proof :-

In ∆ ABC ,

AB = AC

∠ACB= ∠ABC. (opposite angles are equal)

Now again,

In ∆ ACD,

AC= AD

∠ADC = ∠ACD

In ∆ BCD,

∠ABC + ∠BCD + ∠BDC = 180°. (angle sum property of triangle )

∠ACB + ∠BCD + ∠ACD = 180° ( from eq (i) &. (ii)

( ∠ACB + ∠ACD) + (∠BCD) = 180°

( ∠BCD ) + ∠BCD = 180°

2 ∠BCD =180°

∠BCD = 180°/2

∠BCD = 90°

Hence Proved .

Answer 2

$\bf \large \underline{ \sf \: Given : }$

• Triangle ABC such that AB = AC.

• Side BA is produced to D such that AB = AD

$\bf \large \underline{ \sf \: Construction : }$

• Join CD

$\bf \large \underline{ \sf To \: Prove : }$

• Angle BCD = 90°

$\bf \large \underline{ \sf Proof : }$

$\sf \: In \: \triangle \: ABC \: we \: have :$

$\bullet \sf \: AB = AC$

Angles Opposite to equal sides are equal

$\longrightarrow \sf \angle \: ACB = \angle \: ABC....(i)$

Now,

$\bullet \sf \: AB = AD \: \: \: (Given)$

$\therefore \sf \: AD \: = AC$

$\sf \: In \: \triangle \: ADC \: we \: have :$

$\bullet \sf \: AD \: = AC$

Angles Opposite to equal sides are equal

$\longrightarrow \sf \angle \: ACD = \angle \: ADC....(ii)$

After adding equation (i) and equation (ii),we get :

$\sf \angle \: ACB + \angle \: ACD = \angle \: ABC + \angle \: ADC$

Angle ADC is equal to the Angle BDC

$\longrightarrow \sf \angle \: BCD = \angle \: ABC + \angle \: BDC$

Adding Angle BCD on both sides

$\longrightarrow \sf \angle \: BCD + \angle \: BCD = \angle \: ABC + \angle \: BDC + \angle \: BCD$

Sum of the angles of a triangle is 180°

$\longrightarrow \sf 2\angle \: BCD = {180} \degree$

$\longrightarrow \sf \angle \: BCD = \dfrac{180 \degree}{2}$

$\longrightarrow \sf \angle \: BCD = \large \boxed{ \sf \: 90 \degree}$

Hence, Angle BCD is a right angle. Proved!