(i) ∠DAC = ∠BAC ...(1) [∵ Given] ∠DAC = ∠BCA ...(2) [∵ Alternate angle] ∠BAC = ∠ACD ...(3) [∵Alternate angles] From the equations (1), (2) and (3), we have ∠ACD = ∠BCA ...(4) Hence, Diagonal AC bisects angle C also. (ii) From the equation (2) and (4), we have ∠ACD = ∠DAC In ΔADC, ∠ACD = ∠DAC [∵ ProvRead more
(i) ∠DAC = ∠BAC …(1) [∵ Given]
∠DAC = ∠BCA …(2) [∵ Alternate angle]
∠BAC = ∠ACD …(3) [∵Alternate angles]
From the equations (1), (2) and (3), we have
∠ACD = ∠BCA …(4)
Hence, Diagonal AC bisects angle C also.
(ii) From the equation (2) and (4), we have
∠ACD = ∠DAC
In ΔADC,
∠ACD = ∠DAC [∵ Prove above]
AD = DC [∵ In a triangle, the sides opposite to equal angle are equal]
A parallelogram whose adjacent sides are equal, is a rhombus. Hence, ABCD is a rhombus.
(i) In ΔAPD and ΔCQB, DP = BQ [∵ Given] ∠ADP = ∠CBQ [∵ Alternate angle] AD = BC [∵ Opposite sides of a parallelogram] Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule] (ii) ΔAPD ≅ CQB [∵ Prove above] AP = CQ ...(1) [∵ CPCT] (iii) In ΔAQB and ΔCPD, QB = DB [∵ Given] ∠ABQ = ∠CDP [∵ Alternate angle] AB = CD [Read more
(i) In ΔAPD and ΔCQB,
DP = BQ [∵ Given]
∠ADP = ∠CBQ [∵ Alternate angle]
AD = BC [∵ Opposite sides of a parallelogram]
Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule]
(ii) ΔAPD ≅ CQB [∵ Prove above]
AP = CQ …(1) [∵ CPCT]
(iii) In ΔAQB and ΔCPD,
QB = DB [∵ Given]
∠ABQ = ∠CDP [∵ Alternate angle]
AB = CD [∵ Opposite sides of a parallelogram]
Hence, ΔAQB ≅ ΔCPD [ SAS Congruency rule]
(i) In ΔAPB and ΔCQD, ∠APB = ∠CQD [∵ Each 90°] ∠ABP = ∠CDQ [∵ Alternate angles] AB = CD [∵ Opposite sides of a parallelogram] Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule] (ii) ΔAPB ≅ ΔCQD [∵ Prove above] AP = CQ [∵ CPCT]
(i) In ΔAPB and ΔCQD,
∠APB = ∠CQD [∵ Each 90°]
∠ABP = ∠CDQ [∵ Alternate angles]
AB = CD [∵ Opposite sides of a parallelogram]
Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule]
(i) In ABED, AB = DE [∵ Given] AB ∥ DE [∵ Given] Hence, ABED is a parallelogram. (ii) In BEFC, BC = EF [∵ Given] BC ∥ EF [∵ Given] Hence, BEFC is a parallelogram. (iii) In ABED, AD = BE ...(1) [∵ ABED is a parallelogram] AD ∥ BE ...(2) [∵ ABED is a parallelogram] In BEFC, BE = CF ...(3) [∵ ABED is aRead more
(i) In ABED, AB = DE [∵ Given]
AB ∥ DE [∵ Given]
Hence, ABED is a parallelogram.
(ii) In BEFC, BC = EF [∵ Given]
BC ∥ EF [∵ Given]
Hence, BEFC is a parallelogram.
(iii) In ABED,
AD = BE …(1) [∵ ABED is a parallelogram]
AD ∥ BE …(2) [∵ ABED is a parallelogram]
In BEFC,
BE = CF …(3) [∵ ABED is a parallelogram]
BE ∥ CF …(4) [∵ ABED is a parallelogram]
From (2) and (4), we have
AD ∥ CF …(5)
From (1) and (3), we have
AD = CF …(6)
(iv) In ACFD,
AD = CF [∵ From (6)]
AD ∥ CF [∵ From (5)]
Hence, ACFD is a parallelogram
(v) In ACFD,
AC = DF . [∵ ACFD is a parallelogram]
(vi) In ΔABC and ΔDEF
AB = DE [∵ Given]
AC = DF [∵ Proved above]
BC = EF [∵ Given]
Hence, ΔABC ≅ ΔDEF [∵ SSS Congruency rule]
Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD Hence, ar (APB) = (1/2) ar (ABCD) ....(1) [∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.] SiRead more
Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD
Hence, ar (APB) = (1/2) ar (ABCD) ….(1)
[∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.]
Similarly,
Triangle BQC and Parallelogram ABCD lie on the same base AB and between same parallels, AD ∥ BC.
Hence, ar(BQC = (1/2)ar(ABCD) …(2)
From the equation (1) and (2), ar (APB) = ar(BQC).
Diagonal AC of a parallelogram ABCD bisects ∠A see Figure. Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.
(i) ∠DAC = ∠BAC ...(1) [∵ Given] ∠DAC = ∠BCA ...(2) [∵ Alternate angle] ∠BAC = ∠ACD ...(3) [∵Alternate angles] From the equations (1), (2) and (3), we have ∠ACD = ∠BCA ...(4) Hence, Diagonal AC bisects angle C also. (ii) From the equation (2) and (4), we have ∠ACD = ∠DAC In ΔADC, ∠ACD = ∠DAC [∵ ProvRead more
(i) ∠DAC = ∠BAC …(1) [∵ Given]
∠DAC = ∠BCA …(2) [∵ Alternate angle]
∠BAC = ∠ACD …(3) [∵Alternate angles]
From the equations (1), (2) and (3), we have
∠ACD = ∠BCA …(4)
Hence, Diagonal AC bisects angle C also.
(ii) From the equation (2) and (4), we have
∠ACD = ∠DAC
In ΔADC,
∠ACD = ∠DAC [∵ Prove above]
AD = DC [∵ In a triangle, the sides opposite to equal angle are equal]
A parallelogram whose adjacent sides are equal, is a rhombus. Hence, ABCD is a rhombus.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ see Figure. Show that: (i) ∆ APD ≅ ∆ CQB (ii) AP = CQ (iii) ∆ AQB ≅∆ CPD (iv) AQ = CP (v) APCQ is a parallelogram
(i) In ΔAPD and ΔCQB, DP = BQ [∵ Given] ∠ADP = ∠CBQ [∵ Alternate angle] AD = BC [∵ Opposite sides of a parallelogram] Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule] (ii) ΔAPD ≅ CQB [∵ Prove above] AP = CQ ...(1) [∵ CPCT] (iii) In ΔAQB and ΔCPD, QB = DB [∵ Given] ∠ABQ = ∠CDP [∵ Alternate angle] AB = CD [Read more
(i) In ΔAPD and ΔCQB,
DP = BQ [∵ Given]
∠ADP = ∠CBQ [∵ Alternate angle]
AD = BC [∵ Opposite sides of a parallelogram]
Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule]
(ii) ΔAPD ≅ CQB [∵ Prove above]
AP = CQ …(1) [∵ CPCT]
(iii) In ΔAQB and ΔCPD,
QB = DB [∵ Given]
∠ABQ = ∠CDP [∵ Alternate angle]
AB = CD [∵ Opposite sides of a parallelogram]
Hence, ΔAQB ≅ ΔCPD [ SAS Congruency rule]
(iv) ΔAQB ≅ ΔCPD [∵ Prove above]
AQ = CP …(2) [∵ CPCT]
(v) In APCQ,
AP = CQ [∵ From (1)]
AQ = CP [∵ From (2)]
The opposite sides of quadrilateral APCQ are equal.
Hence, APCQ is a parallelogram.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD see Figure. Show that (i) ∆ APB ≅ ∆ CQD (ii) AP = CQ
(i) In ΔAPB and ΔCQD, ∠APB = ∠CQD [∵ Each 90°] ∠ABP = ∠CDQ [∵ Alternate angles] AB = CD [∵ Opposite sides of a parallelogram] Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule] (ii) ΔAPB ≅ ΔCQD [∵ Prove above] AP = CQ [∵ CPCT]
(i) In ΔAPB and ΔCQD,
∠APB = ∠CQD [∵ Each 90°]
∠ABP = ∠CDQ [∵ Alternate angles]
AB = CD [∵ Opposite sides of a parallelogram]
Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule]
(ii) ΔAPB ≅ ΔCQD [∵ Prove above]
AP = CQ [∵ CPCT]
In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). 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) ∆ ABC ≅ ∆ DEF.
(i) In ABED, AB = DE [∵ Given] AB ∥ DE [∵ Given] Hence, ABED is a parallelogram. (ii) In BEFC, BC = EF [∵ Given] BC ∥ EF [∵ Given] Hence, BEFC is a parallelogram. (iii) In ABED, AD = BE ...(1) [∵ ABED is a parallelogram] AD ∥ BE ...(2) [∵ ABED is a parallelogram] In BEFC, BE = CF ...(3) [∵ ABED is aRead more
(i) In ABED, AB = DE [∵ Given]
AB ∥ DE [∵ Given]
Hence, ABED is a parallelogram.
(ii) In BEFC, BC = EF [∵ Given]
BC ∥ EF [∵ Given]
Hence, BEFC is a parallelogram.
(iii) In ABED,
AD = BE …(1) [∵ ABED is a parallelogram]
AD ∥ BE …(2) [∵ ABED is a parallelogram]
In BEFC,
BE = CF …(3) [∵ ABED is a parallelogram]
BE ∥ CF …(4) [∵ ABED is a parallelogram]
From (2) and (4), we have
AD ∥ CF …(5)
From (1) and (3), we have
AD = CF …(6)
(iv) In ACFD,
AD = CF [∵ From (6)]
AD ∥ CF [∵ From (5)]
Hence, ACFD is a parallelogram
(v) In ACFD,
AC = DF . [∵ ACFD is a parallelogram]
(vi) In ΔABC and ΔDEF
AB = DE [∵ Given]
AC = DF [∵ Proved above]
BC = EF [∵ Given]
Hence, ΔABC ≅ ΔDEF [∵ SSS Congruency rule]
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD Hence, ar (APB) = (1/2) ar (ABCD) ....(1) [∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.] SiRead more
Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD
Hence, ar (APB) = (1/2) ar (ABCD) ….(1)
[∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.]
Similarly,
Triangle BQC and Parallelogram ABCD lie on the same base AB and between same parallels, AD ∥ BC.
Hence, ar(BQC = (1/2)ar(ABCD) …(2)
From the equation (1) and (2), ar (APB) = ar(BQC).