In a parellelgram AP =PC prove AP^2+PC^2=DP^2+BP^2
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Answer:
AP BISECTS ∠DAB
∠CPB + ∠DAB =∠APB
Step-by-step explanation:
IN A PARALLELOGRAM ABCD , P IS A POINT ON CD IN SUCH A WAY THAT AD=DP=PC . PROVE THAT AP BISECTS ANGLE DAB
ANGLE CBP+ANGLE DAB= ANGLE APB
lets draw a line PQ ║ AD ║ BC
=> AQ = DP & BQ = PC
given AD = DP = PC
=> AQ = BQ = AD = DP = PC = BC
in ΔADP & ΔAQP
AD = DP = AQ = PQ
& AP is common
=> ΔADP ≅ ΔAQP
=> ∠DAP = ∠QAP
=> ∠DAP = ∠BAP
∠DAB = ∠DAP + ∠BAP
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