    Question

# In the given figure, AB = 6 and BC = 4. BC is extended to E such that C is the midpoint of BE and join AE. It cuts CD in P. Now consider the two points: 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 B Both (i) and (ii) are true Consider △AEB and (\triangle 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=25−16=9 ⇒PC=3 . Since AB = CD (rectangle), CD = 6 and hence DP=CD−PC=6−3=3. That is CD = PC. So P is the midpoint of BC also.  Suggest Corrections  0      Similar questions  Explore more