abcd is a square ef parallel to bd m is the mid point of ef then show that am bisects angle A
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Given in square ABCD, EF||BD
Here diagonal BD bisects ∠B and ∠D.
Hence ∠CBD = ∠ABD = 45° and
∠CDB = ∠ADB = 45°
∴ ∠CBD = ∠CDB = 45°
Given EF||BD and BC is the transversal
Hence ∠CBD = ∠CEF = 45° (Corresponding angles)
Similarly, we get ∠CBD = ∠CFE = 45°
That in ΔCFE, ∠CEF = ∠CFE
⇒ CF = CE → (1)
Consider, BC = CD (Since ABCD is a square)
⇒ CE + BE = CF + DF
⇒ CE + BE = CE + DF [From (1)]
∴ BE = DF
Here diagonal BD bisects ∠B and ∠D.
Hence ∠CBD = ∠ABD = 45° and
∠CDB = ∠ADB = 45°
∴ ∠CBD = ∠CDB = 45°
Given EF||BD and BC is the transversal
Hence ∠CBD = ∠CEF = 45° (Corresponding angles)
Similarly, we get ∠CBD = ∠CFE = 45°
That in ΔCFE, ∠CEF = ∠CFE
⇒ CF = CE → (1)
Consider, BC = CD (Since ABCD is a square)
⇒ CE + BE = CF + DF
⇒ CE + BE = CE + DF [From (1)]
∴ BE = DF
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