In adjacent figure, $∠ B < ∠ A$ and $∠ C < ∠ D$ . Show that $A D < B C$

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Solution

In $△ A O B$ ,

$∠ B < ∠ A$

It is known that in a triangle side opposite to the greater angle is longer.

$∴ A O < B O$ ............. $(1)$

In $△ C O D$ ,

$∠ C < ∠ D \Rightarrow O D < O C$ ........... $(2)$

Add $(1)$ and $(2)$ ,

$\Rightarrow A O + O D < O B + O C$

$∴ A D < B C$

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Similar questions

∠ B < ∠ A

∠ C < ∠ D

A D < B C

A C = A E, A B = A D

∠ B A D = ∠ E A C

B C = D E

Question 3
In the given figure, $∠ B$ < $∠ A$ and $∠ C$ < $∠ D$ . Show that AD < BC.

A D ⊥

B C, B - D - C

{A B}^{2} - {B D}^{2} = {A C}^{2} - {C D}^{2}

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