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UPSC 2017 CDS I Math Paper
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© examsnet.com
Question : 98
Total: 100
In the figure given below, M is the mid-point of AB and ∠DAB = ∠CBA and ∠AMC = ∠BMD. Then the triangle ADM is congruent to the triangle BCM by
[2017 CDS-I]
SAS rule
SSS rule
ASA rule
AAA rule
Validate
Solution:
∠AMC = ∠BMD
Adding ∠CMD to both sides, we have
∴ ∠AMC + ∠CMD
=∠ BMD + ∠CMD
⇒ ∠AMD = ∠BMC
In ΔAMD and ΔBCM
∠DAM = ∠CBM (Given)
AM = BM (M is the mid-point of AB)
∠AMD = ∠BMC (Proved)
∴ ΔAMD ≅ Δ BCM (ASA congruence rule)
© examsnet.com
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