Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that:
(i) it bisects ∠C also
(ii) ABCD is a rhombus.
Given: Diagonal AC of a parallelogram ABCD bisects ∠A.
To Prove: (i) it bisects ∠C also.
(ii) ABCD is a rhombus.
Proof: (i) In ∆ADC and ∆CBA,
AD = CB
| Opp. sides of || gm ABCD
CA = CA | Common
DC = BA
| Opp. sides of || gm ABCD
∴ ∆ADC ≅ ∆CBA
| SSS Congruence Rule
∴ ∠ACD = ∠CAB | C.RC.T.
and ∠DAC = ∠BCA | C.RC.T.
But ∠CAB = ∠DAC | Given
∴ ∠ACD = ∠BCA
∴ AC bisects ∠C also.
(ii) From above,
∠ACD = ∠CAD
∴ AD = CD
| Sides opposite to equal angles of a triangle are equal
∴ AB = BC = CD = DA
| ∵ ABCD is a || gm
∴ ABCD is a rhombus.
(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(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 respectively (see figure). Show that:
(i) ∆APB ≅ ∆CQD
(ii) AP = CQ.