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In the Figure, Bc = Ce and ∠1 = ∠2. Prove that δGcb ≅ δDce.

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Question

In the figure, BC = CE and ∠1 = ∠2. Prove that ΔGCB ≅ ΔDCE.

Sum
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Solution

In ΔGCB and ΔDCE and
∠1 + ∠GBC = ∠2 + ∠DEC = 180°
∠1 = ∠2 = 
⇒ ∠GBC = ∠DEC
BC = CE
∠GCB = ∠DCE =    ...(vertically opposite angles)
Therefore,
ΔGCB ≅ ΔDCE    ....(ASA criteria).

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Chapter 11: Triangles and their congruency - Exercise 11.2

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Frank Mathematics [English] Class 9 ICSE
Chapter 11 Triangles and their congruency
Exercise 11.2 | Q 8

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