Question

ABCD is a rectangle in which diagonal AC bisects angleA as well as angleC. Show that
(i) ABCD is a square
(ii) diagonal BD bisects angleB as well as angleD.​

Answer 1

Solution :

i) Let ABCD be a rectangle such that diagonal AC bisects ∠A as well as ∠C

i.e.,

∠BAC = ∠DAC and ∠BCA = ∠DCA

Since every rectangle is a parallelogram. Therefore,

AB || DC and AC is transversal = ∠BAC = ∠DCA

But, ∠BAC = ∠DAC

∠DAC = ∠DCA

Thus, in ΔADC, we have

∠DAC = ∠DCA

DC = AD ( Sides opposite to equal angles are equal )

But, DC = AB and AD = BC ( ABCD is a rectangle )

AB = BC = CD = DA

Hence, ABCD is a square.

ii) In Δ's BAD and BCD, we have

BA = CD

BC = AD

and, BD = BD [Common]

So, by SSS congruence criterion, we have

ΔBAD = ΔBCD

∠ABD = ∠BCD and ∠ADB = ∠CDB

BD bisects ∠B as well as ∠D.