Question

ABC is an isosceles triangle in which AB=AC, side BA is produced to D such that AD=AB. Show that BCD is right angle??​

Answer 1

Answer:

Given: ∆ABC is an isosceles ∆.

AB = AC and AD = AB

To Prove:

∠BCD is a right angle.

Proof:

In ΔABC,

AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)

In ΔACD,

AD = AB

⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)

Now,

In ΔABC,

∠CAB + ∠ACB + ∠ABC = 180°

⇒ ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — (i)

Similarly in ΔADC,

∠CAD = 180° – 2∠ACD — (ii)

also,

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding (i) and (ii)

∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB – 2∠ACD

⇒ 2∠ACB + 2∠ACD= 360-180

⇒ 2(∠ACB + ∠ACD) = 180°

⇒ ∠BCD = 90°

=========================================

Hope this will help you...

Answer 2

$✧\huge \mathcal \red {\underline{\underline{A}}}\huge \mathcal \green {\underline{\underline{N}}} \huge \mathcal \pink {\underline{\underline{S}}} \huge \mathcal \blue {\underline{\underline{W}}} \huge \mathcal \orange {\underline{\underline {E}}} \huge \mathcal \purple {\underline{\underline{R}}} ✧$

Given :-

• ABC is an isosceles triangle in which AB=AC.
• Side BA is produced to D such that AD=AB.

To prove :-

• /_BCD is right angle.

Proof :-

️ ABC is an isosceles triangle in which AB=AC.

=> /_ABC = /_ACB = x (say)

Now, AD = AB & AB = AC

=>AD = AC

=> /_ACD = /_ADC = y (say)

Now, in ️ BCD

Using angle sum property of triangle

/_BCD + /_DBC + /_BDC = 180°

x + y + y + x = 180°

2(x + y) = 180°

=> x + y = 90°

=> /_BCD = 90°

$\red{\textsf{(✪MARK BRAINLIEST✿)}} \\ \huge\fcolorbox{aqua}{aqua}{\red{FolloW ME}}$

$\huge \fcolorbox{black}{cyan}{♛Hope it helps U♛}$