M is mid point of CD of a parallelogram ABCD. Line BM is drawn intersecting AC at L and AD extended at E. Prove that EL=2BL
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Look at the diagram.
In triangles ΔBMC and ΔDME, angle BMC = angle DME; angle BCM = angle MDE ; and DM = MC. So the two triangles are congruent.
So, BC = DE = AD. => AE = 2 AD = 2 BC
In triangles ΔALE and ΔBLC, angle ALE = angle BLC; angle LAE = LCB ; So the traingles are similar.
AB / BC = 2 => EL / BL = 2 => EL = 2 BL
In triangles ΔBMC and ΔDME, angle BMC = angle DME; angle BCM = angle MDE ; and DM = MC. So the two triangles are congruent.
So, BC = DE = AD. => AE = 2 AD = 2 BC
In triangles ΔALE and ΔBLC, angle ALE = angle BLC; angle LAE = LCB ; So the traingles are similar.
AB / BC = 2 => EL / BL = 2 => EL = 2 BL
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Answer:
Step-by-step explanation:
In triangles ΔBMC and ΔDME, angle BMC = angle DME; angle BCM = angle MDE ; and DM = MC. So the two triangles are congruent.
So, BC = DE = AD. => AE = 2 AD = 2 BC
In triangles ΔALE and ΔBLC, angle ALE = angle BLC; angle LAE = LCB ; So the traingles are similar.
AB / BC = 2 => EL / BL = 2 => EL = 2 BL
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