Question

In ABC, side AB is produced to D such that BD = BC. If /_A = 70° and /_B = 60°.prove that (i) AD > CD (ii) AD > AC.

hope you will help me out!​

Answer 1

ANSWER:

$\sf\green{∠ACB\:=\:50° }$

$\sf\green{Let \:∠BCD\:=\:x,\:∠BDC\:=\:y }$

x+y=60°

$\begin{gathered}\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \bold{\:[exterior\: angle\: property] }} }\\ \\\end{gathered}\end{gathered}$

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$\sf\blue{ Also, \:70\:+\:50\:+\:x+y\:=\:180° }$

$\mathrm{\boxed{\boxed{\pink{→Given \:BD\:=\:BC \: ⇒x=y=30° }}}}$

∠ACD=50° + 30° = 80°

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$\sf\red{ Now, \: in \: {Δ } ^{le} }$

ACD,CD is side opposite to ∠A & AD is side opposite to ∠C.

∵∠C>∠A

∴AD>CD[∵ side opposite to greater angle is longer ]

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Answer 2

Step-by-step explanation:

AD>CD[∵ side opposite to greater angle is longer ]

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