Question

If two angles of a triangle are equal then the sides opposite to them are also equal prove this statement.​

Answer 1

Step-by-step explanation:

Given:

• Two angles of a triangle are equal.

To Prove:

• Sides opposite to equal angles are also equal.

Construction:

• Draw OD , the bisector of ∠A meeting MG at D.

Solution: Let OMG be a triangle in which

• ∠M = ∠G
• OM = OG {to prove}

In ∆OMD and ∆OGD we have

➮ ∠M = ∠G (given)

➮ ∠MOD = ∠GOD (by construction)

➮ OD = OD (common in both ∆s)

∴ ∆OMD = ∆OGD (by Angle Angle Side criteria)

Hence,

• By OM = OG (by CPCT)

$\large\bold{\texttt {Proved }}$

★ See this

• If two opposite angles of a triangle are equal then the side opposite to these angles are also equal. This is a property of isosceles triangle.

Answer 2

$\huge\bf{\underbrace{\gray{TO\:PROVE:-}}}$

• If two angles of a triangle are equal then the sides opposite to them are also equal .

$\huge\bf{\underbrace{\gray{CONSTRUCTION:-}}}$

• Draw a triangle as name ABC .

• Where, ⟨B = ⟨C

• Now, we draw the bisector of ⟨A which meets BC in D .

[NOTE- For Diagram see the attachment.]

$\huge\bf{\underbrace{\gray{PROOF:-}}}$

✞︎ In ∆ABD & ∆ACD, we have

• $\bf\red{\angle{B}\:=\:\angle{C}}$ (Given)

• $\bf\red{\angle{BAD}\:=\:\angle{CAD}}$ ($\because$ AD is bisector of ⟨A)

• $\bf\red{AD\:is\:a\:common\:side\:.}$

$\red\therefore$ By AAS criterion of congruence, we get

• $\bf\purple{\triangle{ABD}\:\cong\:\triangle{ACD}\:}$

$\bf\green{:\implies\:AB\:=\:AC\:}$

☯︎ Hence, from the above it is proved that "If two angles of a triangle are equal then the sides opposite to them are also equal" .