Question

ABCD is parallelogram as shown in the figure BE and DF are perpendiculars on the diagonal Ac. Prove that ∆ BEC ≈ ∆ DFA and BE = DF?​

Answer 1

Answer:

From the figure it is given that,

ABCD is a parallelogram

Perpendiculars DN and BP are drawn on diagonal AC We have to prove that,

(i) ΔDCN≅ΔBAP,

(ii) AN=CP

So, consider the ΔDCN and ΔBAP

AB=DC… [opposite sides of parallelogram are equal]

∠N=∠P... [both angles are equal to 90

∘

]

∠BAP=∠DCN... [alternate angles are equal]

Therefore, ΔDCN≅ΔBAP[AAS axiom]

Then, NC=AP

Because, corresponding parts of congruent triangle.

So, subtracting NP from both sides we get,

NC−NP=AP−NP

AN=CP

Hence it is proved that,

ΔDCN≅ΔBAP and AN=CP.

Step-by-step explanation:

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