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ABCD is a parallelogram. M is the midpoint of BC. Show that DC = CP. [Hint: Prove that ΔABM ≅ ΔPCM] - Mathematics

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Question

ABCD is a parallelogram. M is the midpoint of BC. Show that DC = CP. [Hint: Prove that ΔABM ≅ ΔPCM]

Theorem
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Solution

Given:

ABCD is a parallelogram.

M is the midpoint of BC.

To Prove: DC = CP

Proof:

Step 1: Show that △ABM ≅ △PCM

To prove the congruence of triangles △ABM and △PCM, we need to check for congruence using a valid criterion in this case, SAS (Side-Angle-Side).

1.1: Side 1, AB = PC

In parallelogram ABCD, opposite sides are equal.

Hence, AB = CD and PC = CD.

Thus, AB = PC.

1.2: Side 2, BM = CM

Since M is the midpoint of BC, we have BM = CM.

1.3: Angle, ∠ABM = ∠PCM

In a parallelogram, consecutive angles are supplementary.

So, ∠ABM = ∠PCM.

This is because both are adjacent to the same pair of parallel lines, AB || DC.

Step 2: Apply SAS Congruence

Since:

AB = PC

BM = CM

∠ABM = ∠PCM

We can conclude by the SAS congruence criterion that △ABM ≅ △PCM.

Step 3: DC = CP

From the congruence of triangles △ABM and △PCM, we know that corresponding parts of congruent triangles are equal.

Therefore, DC = CP.

Hence, it is proved that DC = CP.

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Chapter 8: Triangles - MISCELLANEOUS EXERCISE [Page 96]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 8 Triangles
MISCELLANEOUS EXERCISE | Q 10. | Page 96
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