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Question

Refer to the figure shown:

In the given diagram, ABCD is a rectangle. BC is extended to E such that C is the midpoint of BE. The points A and E are joint to meet DC in P. Then

i) P is the midpoint of AE.

ii) P is the midpoint of CD.


A

Both (i) and (ii) are true

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B

Only (i) is true

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C

Only (ii) is true

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D

Neither (i) nor (ii) is true

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Solution

The correct option is A

Both (i) and (ii) are true


Consider AEB and PEC . Since PC is part of DC and DC is parallel to AB,

EAB=EPC [corresponding angles]

AEB=PEC [common angle]

By AA similarity criterion,AEB ~ PEC

So, APPE=BCCE --------------------------(I)

Since C is the mid point of BE, BC=CE. -------------------------(II)

From (II) and (I), APPE = BCCE = 1,

So AP = PE and hence P is the mid point of AE.

Now in right triangle ABE , AB = 6, BE = 2BC = 8.

So by Pythagoras theorem,

AE2=AB2+BE2

AE2=36+64=100

AE=10

Hence PE = AE2 = 102 = 5.

In right triangle PCE, CE = 4 = BC, PE = 5, by Pythagoras theorem,

PE2 = PC2 + CE2

PC2=2516=9

PC=3 .

Since AB = CD (rectangle), CD = 6 and hence DP=CDPC=63=3.

That is, CD = PC. So P is the midpoint of BC also.


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