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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 ΔAPD≅ΔBPC![]() (Hint : In ΔAPD and ΔBPC ¯¯¯¯¯¯¯¯¯AD=¯¯¯¯¯¯¯¯BC,¯¯¯¯¯¯¯¯AP=¯¯¯¯¯¯¯¯BP and ∠PAD=∠PBC=90°−60°=30°] |
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