Question

2. In A ABC. AD is the perpendicular bisector of BC
(see Fig. 5.30). Show that A ABC is an isosceles
triangle in which AB - AC.
Fig. 5.30​

Answer 1

Step-by-step explanation:

Given:

• In ∆ ABC , AD is perpendicular bisector of BC.

To Prove:

• ∆ABC is an isosceles triangle with AB = AC

Solution: We know that perpendicular on any side bisects that side in two equal parts.

Here in ∆ABC we have

• AD ⟂ BC
• BD = DC
• ∠ADB = ∠ADC = 90°

Now we can see that ∆ABC is divided into two triangles ADB and ADC.

In ∆ADB & ∆ADC

➮ AD = AD {common in both ∆s}

➮ BD = DC {from above}

➮ ∠ADB = ∠ADC {from above}

So, ∆ADB ≅ ∆ADC by Side Angle Side Criteria of congruence.

∴ AB = AC {by CPCT}

• If AB = AC then ∠ABC = ∠ACB , angles opposite to equal sides are also equal.

Hence, ∆ABC is an isosceles triangle.

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

Answer 2

Given:

• In ΔABC , AD is the perpendicular bisector of BC.

To Prove:

• ΔABC is an isosceles triangle with AB = AC.

Proof:

As we know,

Perpendicular on any side bisects that side in two equal parts.

So, BD = DC

We can see that, ΔABC is divided into two Δs ADB & ADC.

In ΔADB & ΔADC,

↪ AD = AD [ Common side ]

↪BD = DC [ From above ]

↪ ∠ADB = ∠ADC [ Both angle are 90° ]

↪ .°. ΔADB = ΔADC. [ By SAS test ]

↪ AB = AC [ CPCT ]

↪ ∠ABC = ∠ACB [ Angle opposite to equal sides are equal ]

↪ ΔABC is an isosceles triangle.

Hence Proved!!