Question

In a trapezium ABCD, AB|| CD and AD=BC. If P is the point of intersection of diagonals AC and BD, prove that PA X PC = PB X PD ​

Answer 1

Answer:

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➩ In ΔPDA and ΔPCB

➩ ∠PDA =∠PCB, ∠PAD = ∠PBC

then PA/PB = PD/PC

or PA X PC = PB X PD

➩ ΔPDA~ ΔPCB

______________Or__________________

In △ APB and △ CPD,

➾ ∠APB=∠CPD (Vertically opposite angles)

∠ABP=∠CDP (Alternate angles of parallel sides AB and CD)

∠BAP=∠DCP (Alternate angles of parallel sides AB and CD)

➾ Hence, △APB∼△CPD (AAA rule)

Thus,

➾ PA / PC = PB /PD

➙ PA×PD=PB×PC

$\huge \purple{\fbox {\tt \red{➙PA×PD=PB×PC}}}$

Answer 2

Step-by-step explanation:

In a trapezium ABCD AB|| CD and AD=BC. If P is the point of intersection of diagonals AC and BD prove that PA X PC = PB X PD

In a trapezium ABCD AB|| CD and AD=BC. If P is the point of intersection of diagonals