Question

pls solve this
correct ans and explanation will marked as brainliest​

Answer 1

${\huge{\mathfrak{\blue{\underline{\red{Solution:}}}}}}$

In ∆ABC,

• AC = AB
• $\angle$2 = $\angle$1 ---- (i)

In ∆ADC

• AD = AC
• $\angle$6 = $\angle$5 ---- (ii)

Now,

From exterior angle theoram,

$\pink{\rightarrow}$$\angle$4 = $\angle$1 + $\angle$2

$\pink{\rightarrow}$$\angle$4 = $\angle$2 + $\angle$2 ------- from (i)

$\pink{\rightarrow}$$\angle$4 = 2($\angle$2 ) ----- (iii)

Again,

From exterior angle theoram,

$\pink{\rightarrow}$$\angle$3 = $\angle$6 + $\angle$5

$\pink{\rightarrow}$$\angle$3 = $\angle$5 + $\angle$5 ------- from (ii)

$\pink{\rightarrow}$$\angle$3 = 2($\angle$5 ) ----- (iv)

On adding equation (iii) and equation (iv)

$\pink{\rightarrow}$$\angle$3 + $\angle$4 = 2$\angle$2 + 2$\angle$5

$\pink{\rightarrow}$180° = 2 ( $\angle$2 + $\angle$5) ---- (As 3 and 4 make the linear pair)

$\pink{\rightarrow}$180° = 2 $\angle$BCD

$\pink{\rightarrow}$$\angle$BCD = 90°

Therefore, $\angle$BCD is a right angle.