Question

Prove that angle opposite to equal sides of an isosceles triangle are equal .

Answer 1

Given :

• An isosceles Triangle ABC in which AB = AC .

To Prove :

• ∠B = ∠C

Solution :

Draw a perpendicular AD to BC , AD ⟂ BC .

In Δ ABD and Δ ACD :

$\longmapsto\tt{AB=AC\:(Given)}$

$\longmapsto\tt{\angle{ADB}=\angle{ADC}\:(Each\:90\degree)}$

$\longmapsto\tt{AD=AD\:(Common)}$

So , By SAS Rule Δ ABD ≅ Δ ACD ..

Now ,

$\longmapsto\tt{\angle{B}=\angle{C}\:(By\:CPCT\:Rule)}$

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Triangle :

A close plane figure having three sides and three angles is called as Triangle

Congruent :

Two figures are said to be congruent if they are equal in each aspect. The figures whose shape and size both are same are called Congruent .

CPCT :

CPCT means Corresponding Parts of Congruent Triangle .

SAS Rule :

Two Triangles are congruent if two sides and the included angle of One triangle is equal to the sides and included angle of other Triangle .

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Answer 2

Answer:

Given :-

• ABC is a triangle with AB = AC

To Prove :-

• Angle opposite to AB = Angle opposite to AC. i.e, ∠B = ∠C

Construction :-

• Draw AD perpendicular to BC

$\therefore$ ∠ADB = ∠ADC = 90°

Proof :-

Consider ∆ABD and ∆ACD

where, AD is common

➤ AB = AC

$\therefore$ ∠ADB = ∠ADC = 90°

Hence, ∠ABD = ∠ACD

⇒ ∠ABC = ∠ACB

➠ ∠B = ∠C

$\leadsto\boxed{\bold{\large{PROVED}}}$

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