Short Answer Type
Long Answer TypeShort Answer TypeDiagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that:
(i) it bisects ∠C also
(ii) ABCD is a rhombus.
Given: ABCD is a rhombus.
To Prove: (i) Diagonal AC bisects ∠A as well as ∠C.
(ii) Diagonal BD bisects ∠B as well as ∠D.
Proof: ∵ ABCD is a rhombus
∴ AD = CD
∴ ∠DAC = ∠DCA ...(1)
| Angles opposite to equal sides of a triangle are equal
Also, AD || BC
and transversal AC intersects them
∴ ∠DAC = ∠BCA ...(2)
| Alt. Int. ∠s
From (1) and (2)
∠DCA = ∠BCA
⇒ AC bisects ∠C
Similarly AC bisects ∠A.
(ii) Proceeding similarly as in (i) above, we can prove that BD bisects ∠B as well as ∠D.
Long Answer Type
(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.
Short Answer TypeABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (see figure). Show that:
(i) ∆APB ≅ ∆CQD
(ii) AP = CQ.
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