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Question

In the given figure, BDC=ADC and BCA=ADB. Show that : (i) ACDBDC (ii) BC=AD (iii) A=B.
1346866_3e7137e3e607432fad0a5a10a19a9683.png

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Solution

Let AD and BC intersects at P
Now, in APC & BPD
ACP=BDP (given)
BPD=APC (vertically opposite angles)
So, B=A (angle of triangle)
In ACD & BDC
C=D (Given)
B=A (proved above)
CD=CD (common)
So, ACDBDC (AAS criteria)
So, BC=AD (CPCT)

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