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DENIUS [597]
3 years ago
5

Help Me!

Mathematics
2 answers:
puteri [66]3 years ago
5 0

Answer:

<u>Given:</u>

  • DC ║ AB
  • CM = MB as M is midpoint of BC

i) <u>Since DN and BC are transversals, we have:</u>

  • ∠DCM ≅ ∠NBM and
  • ∠CDM ≅ ∠BNM as alternate interior angles

<u>As two angles and one side is congruent, the triangles are also congruent:</u>

  • ΔDCM ≅ ΔNBM (according to AAC postulate)

So their areas are same.

ii)

<u>The quadrilateral has area of:</u>

  • A(ADCB) = A(ADMB) + A(DCM)

<u>And the triangle has area of:</u>

  • A(ADN) = A(ADMB) + A(NBM)

Since the areas of triangles DCM and NBM are same, the quadrilateral ADCB has same area as triangle ADN.

Sladkaya [172]3 years ago
5 0

Answer:

I think I have proved it before you asked this question before also.

Step-by-step explanation:

SEE the image for solution.

HOPE it helps

Have a great day

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