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:

$\fbox\pink{A}\fbox\green{N}\fbox\pink{S} \fbox\green{W}\fbox\pink{E}\fbox\green{R} \pink{࿐}$.

In ΔPDA and ΔPCB

∠PDA =∠PCB, ∠PAD = ∠PBC

➙then PA/PB = PD/PC

or PA X PC = PB X PD

➙ ΔPDA~ ΔPCB

_______________Or_________________

Step-by-step explanation:

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

${\huge{\boxed{ \mathcal{\orange{Answer !}}}}}$


In ΔPDA and ΔPCB

∠PDA =∠PCB, ∠PAD = ∠PBC

➙then PA/PB = PD/PC

or PA X PC = PB X PD

➙ ΔPDA~ ΔPCB