Question

In a trapezium ABCD , side AB is parallel to side DC ; and the diagonals AC and BD intersect each other at point P . Prove that : (i) triangle APB is similar to triangle CPD . (ii). PA×PD =PB× PC

Answer 1

In triangle APB And Triangle CPD .
AP//DC
AP=PC , DP=PB because digonals intersect each other
So, triangle APB similar to triangle CPD.
PA=PC
PB=PD
PA×PB=PA×PB
So, PA×PD=PC×PB Hence prooved

Answer 2

i). ΔAPB ∼ ΔCPD

ii). PA×PD =PB× PC

Step-by-step explanation:

i). Given that,

In △ APB and △ CPD,

AB ║ DC and

The diagonals AC and BD intersect at P

∵ ∠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       (by AAA similarity rule)

ii). Since △APB ∼ △CPD  using AAA similarity rule

∵ $\frac{PA}{PC}$ = $\frac{PB}{PD}$

Therefore,

PA × PD = PB × PC

Learn more: prove that PA × PD = PB × PC

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