Question

THINK, DISCUSS AND WRITE

Draw two triangles ABC and DBC on the
same base and between the same parallels as shown
in the figure with Pas the point of intersection of AC
and BD. Draw CE || BA and BF || CD such that E
and Flie on line AD.
B.
С
Can you show ar(APAB) = ar(APDC)
Hint: These triangles are not congruent but have equal areas.​

Answer 1

Step-by-step explanation:

Proof In △ABC, we have

PQ∣∣AB

∴

PB

CP

=

QA

CQ

........(i) [By Basic proportionality Theorem]

In △BCD, we have

PR∣∣BD

∴

PB

CP

=

RD

CR

........(ii) [By Thale's Theorem]

From (i) and (ii), we have

QA

CQ

=

RD

CR

Thus, in △ACD, Q and R are points on AC and CD respectively such that

QA

CQ

=

RD

CR

⇒ QR∣∣AD [By the converse of Basic Proportionality Theorem]

Answer 2

Here △ABC and △DBC are on the same base BC and between same parallel lines BC and EF.

Thus, we have, arΔABC=arΔDBC

Now,

arABC=arΔDBC⇒arABC-ar△PBC=arΔDBC-ar△PBC⇒ar(ΔPAB)=ar(ΔPDC).

Hence proved.