Question

In adjacent diagram BF and AE bisect angle ABC and angle BAD respectively of parallelogram ABCD prove that ABCD is a rectangle if AE = EF

Answer 1

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In adjacent diagram BF and AE bisect angle ABC and angle BAD respectively of parallelogram ABCD prove that ABCD is a rectangle if AE = CF

According to question,

∠A = ∠C (Opposite angles)

Line segments AE and CF bisect the

∠A and∠C means,

∠DAE = ∠BCF ----------(i)

Now, In triangles ADE and CBF,

AD = BC (Opposite sides)

∠B = ∠D (Opposite angles)

∠DAE = ∠BCF (from (i))

Therefore, Δ ADE ≅ ΔCBF (By ASA congruency)

DE = BF. (By CPCT)

But, CD = AB

CD - DE = AB - BF.

So, CE = AF.

Therefore, AECF is a quadrilateral having pairs of side parallel and equal,

So, AECF is a parallelogram. Hence, AE || CF