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Quadrilaterals

Question

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

Solution

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Let us join AC.

In ΔABC,

BC = AB (Sides of a rhombus are equal to each other)

∴ ∠1 = ∠2 (Angles opposite to equal sides of a triangle are equal)

However, ∠1 = ∠3 (Alternate interior angles for parallel lines AB and CD)

⇒ ∠2 = ∠3

Therefore, AC bisects ∠C.

Also, ∠2 = ∠4 (Alternate interior angles for || lines BC and DA)

⇒ ∠1 = ∠4

Therefore, AC bisects ∠A.

Similarly, it can be proved that BD bisects ∠B and ∠D as well.

Some More Questions From Quadrilaterals Chapter

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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