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
• 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 (\triangle 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.