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| 1. |
In a quadrilateral, angle A+angle D=90\'. Prove that Ac2+BD2=AD2+BC2 |
| Answer» Given: In quadrilateral ABCD,{tex}\\angle A + \\angle D = 90 ^ { \\circ }{/tex}AC and BD are joinedTo prove: AC2 + BD2 = AD2 + BC2Construction: Produce AB and DC to meet at P.Proof: In\xa0{tex}\\triangle{/tex}APD,{tex}\\angle A + \\angle D = 90 ^ { \\circ }{/tex}(given){tex}\\therefore \\angle P = 90 ^ { \\circ }{/tex}\xa0{tex}\\left( \\because \\angle A + \\angle P + \\angle D = 180 ^ { \\circ } \\right){/tex}Now in right\xa0{tex}\\triangle A C P , \\angle A P D = 90 ^ { \\circ }{/tex}AC2 = PA2 + PC2 ....(i)(Pythagoras Theorem)and in\xa0{tex}\\triangle{/tex}BPDBD2 = PB2 + PD2Adding (i) and (ii)AC2 + BD2 = PA2 + PC2 + PB2 + PD2= (PA2 + PD2) + (PC2 + PB2)= AD2 + BD2({tex}\\therefore{/tex}\xa0In right\xa0{tex}\\triangle{/tex}APD, PA2 + PD2 = AD2 and similarly PC2 + PB2 = BD2) | |