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Home » From the given figure prove that: AP + BQ + CR = BP + CQ + AR. Also, show that AP + BQ + CR = ½ x perimeter of triangle ABC.
From the given figure prove that: AP + BQ + CR = BP + CQ + AR. Also, show that AP + BQ + CR = ½ x perimeter of triangle ABC.
Class 10 | Exercise 18A | ICSE | Maths | Selina | Tangents and Intersecting Chords
From the given figure prove that: AP + BQ + CR = BP + CQ + AR. Also, show that AP + BQ + CR = ½ x perimeter of triangle ABC.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(A) - 11

Solution:

From point

    \[B,\text{ }BQ\text{ }and\text{ }BP\]

are the tangents to the circle

We have,

    \[BQ\text{ }=\text{ }BP\text{ }\ldots \ldots \ldots ..\left( 1 \right)\]

Similarly, we also get

    \[AP\text{ }=\text{ }AR\text{ }\ldots \ldots \ldots \ldots ..\text{ }\left( 2 \right)\]

And,

    \[CR\text{ }=\text{ }CQ\text{ }\ldots \ldots \text{ }\left( 3 \right)\]

Adding

    \[\left( 1 \right),\text{ }\left( 2 \right)\text{ }and\text{ }\left( 3 \right)\]

we get,

    \[AP\text{ }+\text{ }BQ\text{ }+\text{ }CR\text{ }=\text{ }BP\text{ }+\text{ }CQ\text{ }+\text{ }AR\text{ }\ldots \ldots \ldots \text{ }\left( 4 \right)\]

Now, adding

    \[AP\text{ }+\text{ }BQ\text{ }+\text{ }CR\]

to both sides in

    \[\left( 4 \right),\]

we get

    \[2\left( AP\text{ }+\text{ }BQ\text{ }+\text{ }CR \right)\]

    \[=\text{ }AP\text{ }+\text{ }PQ\text{ }+\text{ }CQ\text{ }+\text{ }QB\text{ }+\text{ }AR\text{ }+\text{ }CR\]

    \[2\left( AP\text{ }+\text{ }BQ\text{ }+\text{ }CR \right)\]

    \[=\text{ }AB\text{ }+\text{ }BC\text{ }+\text{ }CA\]

Therefore, we get

    \[AP\text{ }+\text{ }BQ\text{ }+\text{ }CR\]

    \[=~{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }\left( AB\text{ }+\text{ }BC\text{ }+\text{ }CA \right)\]

That is,

    \[AP\text{ }+\text{ }BQ\text{ }+\text{ }CR\]

    \[=~{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }perimeter\text{ }of\text{ }triangle\text{ }ABC\]

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