Question

prove isosceles triangle theorem​

Answer 1

Answer:

Isosceles triangle and proof .

Step-by-step explanation:

Theorem 1) . Angle opposite to the equal side of an isosceles triangle are always equal.

Proof : consider an Isosceles triangle ABC where AC = BC.

we need to prove that theangle opposite to the sideAC and BC are equal , that is,

<CAB = <CBA.

we first draw the bisector of <ACB name it as CD.

Now in triangle ACD and BCD, we have

= AC = BC ( given )

= <ACD = <BCD ( by construction ).

= CD = CD ( common to both )

Thus , ∆ACD congruent ∆BCD ( By SAS )

so, CAB = < CBA ( By C.P.C.T )

Answer 2

Step-by-step explanation:

Theorem 1: Angles opposite to the equal sides of an isosceles triangle are also equal.

Proof: Consider an isosceles triangle ABC where AC = BC.

We need to prove that the angles opposite to the sides AC and BC are equal, that is, ∠CAB = ∠CBA.

Isosceles Triangle

We first draw a bisector of ∠ACB and name it as CD.

Now in ∆ACD and ∆BCD we have,

AC = BC (Given)

∠ACD = ∠BCD (By construction)

CD = CD (Common to both)

Thus, ∆ACD ≅∆BCD (By SAS congruence criterion)

So, ∠CAB = ∠CBA (By CPCT)

Hence proved.

Theorem 2: Sides opposite to the equal angles of a triangle are equal.

Proof: In a triangle ABC, base angles are equal and we need to prove that AC = BC or ∆ABC is an isosceles triangle.

Isosceles Triangle Theorem 2

Construct a bisector CD which meets the side AB at right angles.

Now in ∆ACD and ∆BCD we have,

∠ACD = ∠BCD (By construction)

CD = CD (Common side)

∠ADC = ∠BDC = 90° (By construction)

Thus, ∆ACD ≅ ∆BCD (By ASA congruence criterion)

So, AC = BC (By CPCT)

Or ∆ABC is isosceles.