Q19 of 83 Page 367

In the given figure, BD ‖ CA, E is the midpoint of CA and BD = CA.

Prove that ar(∆ABC) = 2×ar(∆DBC).


Given: BD ‖ CA, E is the midpoint of CA and BD = CA


Consider Δ BCD and Δ DEC


Here,


BD = EC (∵ E is the midpoint of AC that is CE = CA, BD = CA)


CD = CD (common)


∠BDC = ∠ECD (alternate interior angles, DB||AC)


∴ By SAS congruency


Δ BCD Δ DEC


∴ Area (Δ BCD) = Area (Δ DEC) –1


Here,


Area (Δ BCE) = Area (Δ DEC) (triangles on same base CE and between same parallel lines) –2


E is the midpoint of AC, BE is the median of ΔABC


∴ Area (Δ BCE) = Area (Δ ABE) = 1/2 × Area (Δ ABC)


∴ Area (Δ DEC) = 1/2 × Area (Δ ABC) (∵Area (Δ BCE) = Area (Δ DEC))


∴ Area (Δ BCD) = 1/2 × Area (Δ ABC) (∵Area (Δ DEC) = Area (Δ BCD))


More from this chapter

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