InterviewSolution
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InterviewSolution
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In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segment, AF and EC trisect the diagonal BD |
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Answer» Given: ABCD is a parallelogram in which E and F are the mid-points of sides AB and CD respectively. To prove: BQ = QP = PD Proof: ABCD is a parallelogram ⇒ AB || DC ⇒ AE || AF and AE = CF (∵ \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) CD) ⇒ AECF is a parallelogram. ⇒ EC || AF …(i) If ΔAPB, E is mid-point of AB (given) and EQ ||AP [using (ii)] ⇒ Q is the mid-point PB ⇒ BQ = PQ …(ii) Similarly, in ΔDQC, we can show that DP = PQ From (ii) and (iii), we have BQ = PQ = PD Hence, line segment AF and EC trisect the diagonal BD. |
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