Question

prove that triangle ABC is isosceles if any one of following holds
(1) altitude AD bisects
(2) median AD is perpendicular to the base BC

Answer 1

Correct question:-

Prove that triangle ABC is isosceles if any one of following holds

(1) altitude AD bisects BC

(2) median AD is perpendicular to the base BC.

Solution :-

Given that:-

• Altitude AD bisects BC.
• AD ⊥ BC

To Prove:-

∆ABC is an isosceles triangle.

Proof:-

In ∆ABD and ∆ACD,

• AD = AD [ ∵ common]
• ∠ADB = ∠ADC [ ∵ AD ⊥ BC]
• BD = CD [AD bisects BC]

Therefore, ∆ABD ≈ ∆ACD (SAS criteria)

And, AB = AC (by CPCT)

Now,

In ∆ABC,

As we know that AB = AC.

• And if two sides of triangle are equal, then it's considered as an isosceles triangle.

So, ∆ABC is an isosceles triangle.

Hence Proved!

Answer 2

Question :-

Prove that triangle ABC is isosceles if any one of following holds :-

• altitude AD bisects BC
• median AD is perpendicular to the base BC

Solution:-

Given:-

• Altitude AD bisects BC

• Median AD is perpendicular to the base BC

To prove :-

• Triangle ABC is isosceles

Proof :-

In triangle ABD and triangle ACD, we have

• BD = DC [ As Altitude AD bisects BC ]

• Angle ADB = Angle ACD [ Given , AD ⊥ BC]

• AD is common side

.°. ∆ABD ≅ ∆ACD [By S-A-S congruence rule]

.°. AB = AC [ CPCT ]

We know that ,

When two sides of triangle are of equal length, then triangle is called Isosceles. Two angles of isosceles triangle are also equal.

.°. ABC is an isosceles triangle [Hence Proved]

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