Question

The following figure shows a trapezium ABCD in which AB||DC. p is yhe mid point of AD and PR || AB. Prove that PR= ½(AB + CD)

Answer 1

Given ABCD is a trapezium in which AB || DC and EF||AB|| CD.
Construction Join, the diagonal AC which intersects EF at O.
To show F is the mid-point of BC. 



Proof Now, in ΔADC, E is the mid-point of AD and OE || CD. Thus, by mid-point theorem, O is mid-point of AC.
Now, in ΔCBA, 0 is the mid-point of AC and OF || AB.
So, by mid-point theorem, F is the mid-point of BC.

Answer 2

Qᴜᴇsᴛɪᴏɴ -

The following figure shows a trapezium ABCD in which AB || DC. P is the midpoint of AD and PR || AB. Prove that :

$\sf \: PR = \dfrac{1}{2} (AB + CD)$

ɢɪᴠᴇɴ -

• AB || DC
• PR || AB
• P is the midpoint of AD.

ᴛᴏ ᴘʀᴏᴠᴇ -

$\sf \: PR = \dfrac{1}{2} (AB + CD)$

ᴘʀᴏᴏғ -

In ∆ ABD,

PQ = ½AB ....(i) [mid point theorem]

In ∆ BCD,

Q is a midpoint of BD. [Conv. mid point theorem]

➠ QR = 1/2DC ....(ii) [mid point theorem]

On adding eq. (i) & (ii)

$\bold{ \implies \: PQ + QR = \dfrac{1}{2} AB + \dfrac{1}{2} DC} \\ \\ \bold{ \implies \: PQ + QR = \dfrac{1}{2} (AB + DC)}\\ \\ \green{ \sf \: ∴ \:PR = \dfrac{1}{2} (AB + DC) \: \: \: \: \: \: \: \: \: \: \: \red{(proved)} }$

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@MissSolitary ✌️

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