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In the adjacent figure ABCD is a square and ΔAPB is an equilateral triangle. Prove that ΔAPD≅ΔBPC(Hint : In ΔAPD and ΔBPC ¯¯¯¯¯¯¯¯¯AD=¯¯¯¯¯¯¯¯BC,¯¯¯¯¯¯¯¯AP=¯¯¯¯¯¯¯¯BP and ∠PAD=∠PBC=90°−60°=30°]

Answer» In the adjacent figure ABCD is a square and ΔAPB is an equilateral triangle. Prove that ΔAPDΔBPC



(Hint : In ΔAPD and ΔBPC ¯¯¯¯¯¯¯¯¯AD=¯¯¯¯¯¯¯¯BC,¯¯¯¯¯¯¯¯AP=¯¯¯¯¯¯¯¯BP and PAD=PBC=90°60°=30°]


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