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Quadrilaterals

Question

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Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). Show that
(i) It bisects ∠C also,
(ii) ABCD is a rhombus.

Solution

(i) ABCD is a parallelogram.

∴ ∠DAC = ∠BCA (Alternate interior angles) … (1)

And, ∠BAC = ∠DCA (Alternate interior angles) … (2)

However, it is given that AC bisects ∠A.

∴ ∠DAC = ∠BAC … (3)

From equations (1), (2), and (3), we obtain

∠DAC = ∠BCA = ∠BAC = ∠DCA … (4)

⇒ ∠DCA = ∠BCA

Hence, AC bisects ∠C.

(ii)From equation (4), we obtain

∠DAC = ∠DCA

∴ DA = DC (Side opposite to equal angles are equal)

However, DA = BC and AB = CD (Opposite sides of a parallelogram)

∴ AB = BC = CD = DA

Hence, ABCD is a rhombus.

Some More Questions From Quadrilaterals Chapter

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). Show that
(i) It bisects ∠C also,
(ii) ABCD is a rhombus.

ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure).
Show that:
(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the given figure). Show that
(i) Quadrilateral ABED is a parallelogram
(ii) Quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) Quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF.

ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) diagonal AC = diagonal BD

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