Question

Show that angles opposite to equal sides of an isosceles triangle are equal. Apply this result to show that the angles of an equilateral triangle are 60 degree each...Urgently required

Answer 1

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:

Answer 2

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