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
• B. Only (i) is true
• C. Only (ii) is true
• D. Neither (i) nor (ii) is true

Answer 1

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,

$\angle EAB=\angle EPC$ [corresponding angles]

$\angle AEB=\angle PEC$ [common angle]

By AA similarity criterion,$△AEB$ ~ $△PEC$

So, $\frac{AP}{PE}=\frac{BC}{CE}$ --------------------------(I)

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

From (II) and (I), $\frac{AP}{PE}$ = $\frac{BC}{CE}$ = 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,

$\Rightarrow A{E}^2=A{B}^2+B{E}^2$

$\Rightarrow A{E}^2=36+64=100$

$\Rightarrow AE=10$

Hence PE = $\frac{AE}{2}$ = $\frac{10}{2}$ = 5.

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

${PE}^2$ = ${PC}^2$ + ${CE}^2$

$\Rightarrow$ ${PC}^2=25-16=9$

$\Rightarrow PC=3$ .

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.