ABCD is a quadrilateral. A line through D, parallel to AC meets BC produced in P. Prove that ar(triangle APD)= ar(quad. ABCD)

Asked by Sthitaprajna Mishra | 4th Feb, 2014, 08:13: PM

Expert Answer:

 
Consider the above figure.
 
Two triangles with same base and between the same parallels are of equal areas.
 
Thus, a r open parentheses triangle A C D close parentheses equals a r open parentheses triangle A C P close parentheses
Now subtracting the area of triangle AOC from both the sides of the equation, we have
 
rightwards double arrow a r open parentheses triangle A C D close parentheses minus a r open parentheses triangle A O C close parentheses equals a r open parentheses triangle A C P close parentheses minus a r open parentheses triangle A O C close parentheses rightwards double arrow a r open parentheses triangle A O D close parentheses equals a r open parentheses triangle C O P close parentheses N o w space a d d space t h e space a r e a space o f space t h e space q u a d r i l a t e r a l space A B C O space t o space b o t h space t h e space s i d e s comma space w e space h a v e rightwards double arrow a r open parentheses triangle A O D close parentheses plus a r open parentheses square A B C O close parentheses equals a r open parentheses triangle C O P close parentheses plus a r open parentheses square A B C O close parentheses rightwards double arrow a r open parentheses square A B C D close parentheses equals a r open parentheses triangle A B P close parentheses
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Answered by  | 6th Feb, 2014, 11:25: AM

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