Given: An Isosceles Δ ABC in which AB=AC

To Prove: ∠B=∠C

Construction: Draw AM⊥BC

Proof: In right Δ AMB and Δ A MC

→AB= AC   [given]

→∠A MB = ∠A MC [Each being 90°]

→AM is common.

Δ AMB ≅ Δ A MC [ R HS congruency,→ which states if in a right angled triangle hypotenuse and one side is equal to hypotenuse and other side then two triangles are congruent.]

∠B=∠C [C PCT]

Now, If all sides of a triangle are equal , then

∠A=∠B=∠C

As ,∠A+∠B+∠C=180° [By angle sum property of quadrilateral]

3×∠A=180°

∠A=180°÷3=60°

Which shows, ∠A=∠B=∠C=60°, i.e triangle is equilateral.

Proof:

Step-by-step explanation:

ANSWER

ABC is a given triangle with, AB=AC.

To prove: Angle opposite to AB= Angle

opposite to AC (i.e) ∠C=∠B

Construction: Draw AD perpendicular to BC

∴∠ADB=∠ADC=90

o

Proof:

Consider △ABD and △ACD

AD is common

AB=AC

∴∠ADB=∠ADC=90

o

Hence ∠ABD=∠ACD

∠ABC=∠ACB

∠B=∠C. Hence the proof

This is known as Isosceles triangle theorem