In the given figure- D is a point on side BC of a- ABC-and E is a point such that CD - DE-Prove that AB - AC-BE-

Given: CD = DETo prove: AB + AC gt; BE Proof: In ∆ABC, AB+AC gt;BC— 1 In ∆BED, BD+CD gt;BE⇒BC gt;BE— 2 From (1) and (2), we get AB + AC gt; BE ...

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Standard IX

Mathematics

Relationship between Unequal Sides of Triangle and the Angles Opposite to It.

In the given ...

Question

$In the given figure, D is a point on side BC of a Δ A B C and E is a point such that CD = DE. Prove that AB + AC > B E .$

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Solution

Given: CD = DETo prove: AB + AC > BE
Proof:
In ∆ABC,
AB+AC>BC----1
In ∆BED,
BD+CD>BE⇒BC>BE----2
From (1) and (2), we get
AB + AC > BE

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In the given figure, D is a point on side BC of a ΔABC and E is a point such that CD = DE.Prove that AB + AC>BE.

Given: CD = DETo prove: AB + AC > BE Proof: In ∆ABC, AB+AC>BC----1 In ∆BED, BD+CD>BE⇒BC>BE----2 From (1) and (2), we get AB + AC > BE

$In the given figure, D is a point on side BC of a Δ A B C and E is a point such that CD = DE. Prove that AB + AC > B E .$

Given: CD = DETo prove: AB + AC > BE
Proof:
In ∆ABC,
AB+AC>BC----1
In ∆BED,
BD+CD>BE⇒BC>BE----2
From (1) and (2), we get
AB + AC > BE