Question

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(∆APB)×ar(∆CPD) =ar (∆APD)×ar(∆BPC)
​

Answer 1

Given :

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.

To Prove :

$\displaystyle{\sf{ar(∆APB) \times ar(∆CPD) =ar (∆APD) \times ar(∆BPC)}}$

Construction :

From A to C , draw perpendiculars AE and CF respectively to BD.

Proof :

$\displaystyle{\sf{ar(∆APB) \times ar(∆CPD)}}$

:$\Rightarrow{\sf{ \dfrac{(PB)(AE)}{2} \times \Bigg[ \dfrac{DP \times CF}{2} \Bigg]}}$

$\displaystyle{\sf{ \left| \bigg( Area\:of\:a\:∆\:= \frac{Base \times Corresponding\:altitude}{2} \bigg)}}$

:$\Rightarrow{\sf{ \dfrac{1}{4}(PB)(AE)(DP)(CF)\:\:\:\:\:......(1)}}$

$\displaystyle{\sf{ar(∆APD) \times ar(∆BPC)}}$

:$\Rightarrow{\sf{ \dfrac{(DP)(AE)}{2} \times \Bigg[ \dfrac{PB \times CF}{2} \Bigg]}}$

:$\Rightarrow{\sf{ \dfrac{1}{4}(PB)(AE)(DP)(CF)\:\:\:\:\:......(2)}}$

From (1) and (2),

${\boxed{\sf{\red{ar(∆APB) \times ar(∆CPD) =ar (∆APD) \times ar(∆BPC)}}}}$

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Answer 2

$\large\underline\bold{ANSWER \huge{\red{\checkmark}}}$

$\large\underline\bold{GIVEN,}$

$\dashrightarrow ABCD\:is\:a\: quadrilateral\\ \dashrightarrow AC\:and\:BD\:are\:diagonals\\which\: interests\:at\:point\:p\\ \dashrightarrow let\:the\:height\:of\:\triangle\: APD\:be\:‘x’\\ \dashrightarrow and\:of\: \triangle BPC\:be\:y.$

$\large\underline\bold{TO\:PROVE,}$

$\dashrightarrow \large{ar(\triangle APB )\times ar \: (\triangle CPD) = ar(\triangle BPC ) \times ar (\triangle APD ) }$

✯FORMULA IN USE,

$\large{\boxed{\bf{\star\:\: area\:of\:\triangle= \dfrac{1}{2} \times b\times h\:\:\star }}}$

$\huge\underline\mathrm{\red{PROOF,}}$

$\therefore using\:formula,$

In ∆APD and ∆ BPC

$\dashrightarrow Ar(\triangle APD)= \dfrac{1}{2}\times x\times DP\:---\boxed{1}$

$\dashrightarrow Ar(\triangle BPC)= \dfrac{1}{2}\times y\times BP\:--\boxed{2}$

Now, in ∆ APD and ∆APB,

$\dashrightarrow Ar(\triangle APB)= \dfrac{1}{2}\times x\times BP\:--\boxed{3}$

$\dashrightarrow Ar(\triangle APD)= \dfrac{1}{2}\times y\times DP\:--\boxed{4}$

From eq [1], and [4]

$\dashrightarrow we \:get\: \\ in\:\triangle APD \: \\ x=y \:--\boxed{5}$

From eq[5] .We get,

$\therefore \triangle BPC= \triangle APB\:---\boxed{6}$

From eq [1],[2],[3],[4],[5],[6]. We get,

$\rm{\boxed{\bf{ \star\:\:ar(\triangle APB )\times ar \: (\triangle CPD) = ar(\triangle BPC ) \times ar (\triangle APD ) \:\: \star}}}$

$\large\underline\bold{HENCE\:PROOVED\huge{\purple{\checkmark}},}$

$\pink{\text{NOTE:- FOR DIAGRAM REFER THE ATTACHMENT.}}$

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