Question

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

Answer 1

Answer: ar(∆ABP) = ar(ABCD)

Step-by-step explanation:

Given : ABCD is quad where line parallel to AC through D meets at BC produced on P.

To Prove : ar(∆ABP) = ar(ABCD)

Proof : Since AC || DP therefore,

ar(∆ADC) = ar(∆ACP)

Because area of triangles at same base and parallels are equal.

Adding ar(∆ABC) both sides we get,

ar(∆ADC) + ar(∆ABC) = ar(∆ACP) + ar(∆ABC)

ar(ABCD) = ar(∆APB)

OR

ar(∆ABP) = ar(ABCD)

Q.E.D

Answer 2

Solu:- Given:- ABCD is quadrilateral A line through D parallel to AC meets BC produced in P.

Prove that:- ar (ABP) = ar (quad ABCD)

Proof:- Since, In ∆APC and ∆ADC are lie on the same base AC and between the same parallel line AD and CP.

So, ar (APC) = ar (ADC)
By adding both sides ∆ABC ------ (1)
so, we get , ar (APC) + ar (ABC)=ar (ADC) + ar (ABC)
ar (ABP) = ar (quad ABCD)

Hope it is helpful to you....