# Illustrative Examples Based on Inequality a Triangle

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In Î”ABC, AB > AC and D is a point on BC. Show thatÂ AB > AD.

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**Given: ** In Î”ABC, AB > AC and D is a point on BC.

**To Prove:** AB > AD.

**Proof:**

Since in Î”ABC,

AB > AC

*. ^{.}. * âˆ ACB > âˆ ABC ...(i)

Now in Î”ADC, âˆ ADB is the exterior angle of Î”ADC.

âˆ´

*âˆ ADB = âˆ DAC + âˆ ACD*

âˆ´

*âˆ ADB > âˆ ACD ... (ii)*

From inequality (i) and (ii), we getÂ âˆ ADB > âˆ ACD > âˆ ABC

âˆ´

*Â*âˆ ADB > âˆ ABC

âˆ´Â âˆ ADB > âˆ ABD

âˆ´

*Â*AB > AD ...[side opposite to greater angle]

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In the interior of Î”PQR, S is any point, show that PQ + PR > SQ + SR.

**Given:** In the interior of Î”PQR, S is any point.

**To Prove:** PQ + PR > SQ + SR

**Construction:** Produce QS to meet PR at T.

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**Proof:**

In Î”PQT, we have

PQ + PT > QT ... [sum of two sides greater than third]

âˆ´ PQ + PT > QS + ST ... (i)

Now in Î”RST, we have

ST + TR > RS ... (ii)

Adding inequalities (i) and (ii), we get PQ + PT + ST + TR > QS + ST + RS

âˆ´* *PQ + (PT + TR) > QS + RS

âˆ´* *PQ + PR > QS + RS

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In a quadrilateral PQRS, PQ is its longest side and RS is its shortest side. Prove that âˆ R > âˆ P andÂ âˆ S > âˆ Q.

**Given: **In a quadrilateral PQRS, PQ is its longest side and RS is its shortest side.

**To Prove :** âˆ R > âˆ P and âˆ S > âˆ Q.

**Construction: ** Join PR and QS

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**Proof: **

In Î”PRS, PS > SR

âˆ´*Â Â Â Â Â Â *âˆ 3 > âˆ 1 ...... (i)

Now in Î”PQR, PQ > QR

âˆ´*Â Â Â Â Â Â Â *âˆ 4 > âˆ 2 ...... (ii)

Adding (i) and (ii), we get

Â Â Â Â Â Â Â Â Â Â Â âˆ 3 + âˆ 4 > âˆ 1 + âˆ 2

âˆ´*Â Â Â Â Â Â Â *âˆ R > âˆ P

Similarly, we can prove that

Â Â Â Â Â Â Â Â âˆ S > âˆ Q